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Title: Calculus Made Easy
Being a very-simplest introduction to those beautiful
methods which are generally called by the terrifying names
of the Differentia
Author: Silvanus Thompson
Release Date: October 9, 2012 [EBook #33283]
Language: English
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CALCULUS MADE EASY
MACMILLAN AND CO., Limited
LONDON : BOMBAY : CALCUTTA
MELBOURNE
THE MACMILLAN COMPANY
NEW YORK : BOSTON : CHICAGO
DALLAS : SAN FRANCISCO
THE MACMILLAN CO. OF CANADA, Ltd.
TORONTO
CALCULUS MADE EASY:
BEING A VERY-SIMPLEST INTRODUCTION TO
THOSE BEAUTIFUL METHODS OF RECKONING
WHICH ARE GENERALLY CALLED BY THE
TERRIFYING NAMES OF THE
DIFFERENTIAL CALCULUS
AND THE
INTEGRAL CALCULUS.
BY
F. R. S.
SECOND EDITION, ENLARGED
MACMILLAN AND CO., LIMITED
ST. MARTIN’S STREET, LONDON

COPYRIGHT.
First Edition 1910.
Reprinted 1911 (twice), 1912, 1913.
Second Edition 1914.
What one fool can do, another can.
(Ancient Simian Proverb.)
PREFACE TO THE SECOND EDITION.
The surprising success of this work has led the author to add a considerable number of worked examples and exercises. Advantage has
also been taken to enlarge certain parts where experience showed that
further explanations would be useful.
The author acknowledges with gratitude many valuable suggestions
and letters received from teachers, students, and—critics.
October, 1914.
CONTENTS.
Chapter
Page
Prologue . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
ix
I. To deliver you from the Preliminary Terrors
1
II. On Different Degrees of Smallness . . . . . . . . . . .
3
III. On Relative Growings . . . . . . . . . . . . . . . . . . . . . . . . . .
9
IV. Simplest Cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17
V. Next Stage. What to do with Constants . . . . . .
25
VI. Sums, Differences, Products and Quotients . . .
34
VII. Successive Differentiation . . . . . . . . . . . . . . . . . . . . .
48
VIII. When Time Varies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
52
IX. Introducing a Useful Dodge . . . . . . . . . . . . . . . . . . .
66
X. Geometrical Meaning of Differentiation . . . . . .
75
XI. Maxima and Minima . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
91
XII. Curvature of Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . 109
XIII. Other Useful Dodges . . . . . . . . . . . . . . . . . . . . . . . . . . 118
XIV. On true Compound Interest and the Law of Organic Growth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
vii
CALCULUS MADE EASY
Chapter
viii
Page
XV. How to deal with Sines and Cosines . . . . . . . . . . . 162
XVI. Partial Differentiation . . . . . . . . . . . . . . . . . . . . . . . . 172
XVII. Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180
XVIII. Integrating as the Reverse of Differentiating 189
XIX. On Finding Areas by Integrating . . . . . . . . . . . . . . 204
XX. Dodges, Pitfalls, and Triumphs . . . . . . . . . . . . . . . . 224
XXI. Finding some Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 232
Table of Standard Forms . . . . . . . . . . . . . . . . . . . . . . . . 249
Answers to Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . 252
PROLOGUE.
Considering how many fools can calculate, it is surprising that it
should be thought either a difficult or a tedious task for any other fool
to learn how to master the same tricks.
Some calculus-tricks are quite easy. Some are enormously difficult.
The fools who write the textbooks of advanced mathematics—and they
are mostly clever fools—seldom take the trouble to show you how easy
the easy calculations are. On the contrary, they seem to desire to
impress you with their tremendous cleverness by going about it in the
most difficult way.
Being myself a remarkably stupid fellow, I have had to unteach
myself the difficulties, and now beg to present to my fellow fools the
parts that are not hard. Master these thoroughly, and the rest will
follow. What one fool can do, another can.
CHAPTER I.
TO DELIVER YOU FROM THE PRELIMINARY
TERRORS.
The preliminary terror, which chokes off most fifth-form boys from
even attempting to learn how to calculate, can be abolished once for
all by simply stating what is the meaning—in common-sense terms—of
the two principal symbols that are used in calculating.
These dreadful symbols are:
(1) d which merely means “a little bit of.”
Thus dx means a little bit of x; or du means a little bit of u. Ordinary mathematicians think it more polite to say “an element of,”
instead of “a little bit of.” Just as you please. But you will find that
these little
Z bits (or elements) may be considered to be indefinitely small.
(2)
which is merely a long S, and may be called (if you like) “the
sum of.” Z
Z
Thus dx means the sum of all the little bits of x; or dt means
the sum of all the little bits of t. Ordinary mathematicians call this
symbol “the integral of.” Now any fool can see that if x is considered
as made up of a lot of little bits, each of which is called dx, if you
add them all up together you get the sum of all the dx’s, (which is the
CALCULUS MADE EASY
2
same thing as the whole of x). The word “integral” simply means “the
whole.” If you think of the duration of time for one hour, you may (if
you like) think of it as cut up into 3600 little bits called seconds. The
whole of the 3600 little bits added up together make one hour.
When you see an expression that begins with this terrifying symbol, you will henceforth know that it is put there merely to give you
instructions that you are now to perform the operation (if you can) of
totalling up all the little bits that are indicated by the symbols that
follow.
That’s all.
CHAPTER II.
ON DIFFERENT DEGREES OF SMALLNESS.
We shall find that in our processes of calculation we have to deal with
small quantities of various degrees of smallness.
We shall have also to learn under what circumstances we may consider small quantities to be so minute that we may omit them from
consideration. Everything depends upon relative minuteness.
Before we fix any rules let us think of some familiar cases. There
are 60 minutes in the hour, 24 hours in the day, 7 days in the week.
There are therefore 1440 minutes in the day and 10080 minutes in the
week.
Obviously 1 minute is a very small quantity of time compared with
a whole week. Indeed, our forefathers considered it small as compared with an hour, and called it “one minùte,” meaning a minute
fraction—namely one sixtieth—of an hour. When they came to require still smaller subdivisions of time, they divided each minute into
60 still smaller parts, which, in Queen Elizabeth’s days, they called
“second minùtes” (i.e. small quantities of the second order of minuteness). Nowadays we call these small quantities of the second order of
smallness “seconds.” But few people know why they are so called.
Now if one minute is so small as compared with a whole day, how
4
CALCULUS MADE EASY
much smaller by comparison is one second!
Again, think of a farthing as compared with a sovereign: it is barely
worth more than
1
1000
part. A farthing more or less is of precious little
importance compared with a sovereign: it may certainly be regarded
as a small quantity. But compare a farthing with £1000: relatively to
this greater sum, the farthing is of no more importance than
1
1000
of a
farthing would be to a sovereign. Even a golden sovereign is relatively
a negligible quantity in the wealth of a millionaire.
Now if we fix upon any numerical fraction as constituting the proportion which for any purpose we call relatively small, we can easily
state other fractions of a higher degree of smallness. Thus if, for the
purpose of time,
1
60
be called a small fraction, then
1
60
of
1
60
(being a
small fraction of a small fraction) may be regarded as a small quantity
of the second order of smallness.∗
Or, if for any purpose we were to take 1 per cent. (i.e.
1
)
100
as a
1
) would be a
10,000
1
and 1,000,000
would be
small fraction, then 1 per cent. of 1 per cent. (i.e.
small fraction of the second order of smallness;
a small fraction of the third order of smallness, being 1 per cent. of
1 per cent. of 1 per cent.
Lastly, suppose that for some very precise purpose we should regard
1
1,000,000
as “small.” Thus, if a first-rate chronometer is not to lose
or gain more than half a minute in a year, it must keep time with
an accuracy of 1 part in 1, 051, 200. Now if, for such a purpose, we
∗
The mathematicians talk about the second order of “magnitude” (i.e. great-
ness) when they really mean second order of smallness. This is very confusing to
beginners.
5
DIFFERENT DEGREES OF SMALLNESS
regard
1
1,000,000
1
,
1,000,000
that is
(or one millionth) as a small quantity, then
1
1,000,000,000,000
1
1,000,000
of
(or one billionth) will be a small quantity
of the second order of smallness, and may be utterly disregarded, by
comparison.
Then we see that the smaller a small quantity itself is, the more
negligible does the corresponding small quantity of the second order
become. Hence we know that in all cases we are justified in neglecting
the small quantities of the second—or third (or higher)—orders, if only
we take the small quantity of the first order small enough in itself.
But, it must be remembered, that small quantities if they occur in
our expressions as factors multiplied by some other factor, may become
important if the other factor is itself large. Even a farthing becomes
important if only it is multiplied by a few hundred.
Now in the calculus we write dx for a little bit of x. These things
such as dx, and du, and dy, are called “differentials,” the differential
of x, or of u, or of y, as the case may be. [You read them as dee-eks,
or dee-you, or dee-wy.] If dx be a small bit of x, and relatively small of
itself, it does not follow that such quantities as x · dx, or x2 dx, or ax dx
are negligible. But dx × dx would be negligible, being a small quantity
of the second order.
A very simple example will serve as illustration.
Let us think of x as a quantity that can grow by a small amount so
as to become x + dx, where dx is the small increment added by growth.
The square of this is x2 + 2x · dx + (dx)2 . The second term is not
negligible because it is a first-order quantity; while the third term is of
the second order of smallness, being a bit of, a bit of x2 . Thus if we
6
CALCULUS MADE EASY
1
60
took dx to mean numerically, say,
be
2
60
of x, then the second term would
of x2 , whereas the third term would be
1
3600
of x2 . This last term
is clearly less important than the second. But if we go further and take
dx to mean only
1
1000
of x, then the second term will be
the third term will be only
1
1,000,000
2
1000
of x2 , while
of x2 .
x
x
Fig. 1.
Geometrically this may be depicted as follows: Draw a square
(Fig. 1) the side of which we will take to represent x. Now suppose
the square to grow by having a bit dx added to its size each way.
The enlarged square is made up of the original square x2 , the two
rectangles at the top and on the right, each of which is of area x · dx
(or together 2x · dx), and the little square at the top right-hand corner
which is (dx)2 . In Fig. 2 we have taken dx as quite a big fraction
of x—about
1
.
5
But suppose we had taken it only
1
—about
100
the
thickness of an inked line drawn with a fine pen. Then the little corner
square will have an area of only
1
10,000
of x2 , and be practically invisible.
Clearly (dx)2 is negligible if only we consider the increment dx to be
itself small enough.
Let us consider a simile.
7
DIFFERENT DEGREES OF SMALLNESS
x
dx
dx
dx
x
x
x
x · dx
(dx)2
x2
x · dx
dx
Fig. 2.
Fig. 3.
Suppose a millionaire were to say to his secretary: next week I will
give you a small fraction of any money that comes in to me. Suppose
that the secretary were to say to his boy: I will give you a small fraction
of what I get. Suppose the fraction in each case to be
1
100
part. Now
if Mr. Millionaire received during the next week £1000, the secretary
would receive £10 and the boy 2 shillings. Ten pounds would be a
small quantity compared with £1000; but two shillings is a small small
quantity indeed, of a very secondary order. But what would be the
disproportion if the fraction, instead of being
1
1000
1
,
100
had been settled at
part? Then, while Mr. Millionaire got his £1000, Mr. Secretary
would get only £1, and the boy less than one farthing!
The witty Dean Swift∗ once wrote:
“So, Nat’ralists observe, a Flea
“Hath smaller Fleas that on him prey.
“And these have smaller Fleas to bite ’em,
“And so proceed ad infinitum.”
∗
On Poetry: a Rhapsody (p. 20), printed 1733—usually misquoted.
CALCULUS MADE EASY
8
An ox might worry about a flea of ordinary size—a small creature of
the first order of smallness. But he would probably not trouble himself
about a flea’s flea; being of the second order of smallness, it would be
negligible. Even a gross of fleas’ fleas would not be of much account to
the ox.
CHAPTER III.
ON RELATIVE GROWINGS.
All through the calculus we are dealing with quantities that are growing, and with rates of growth. We classify all quantities into two classes:
constants and variables. Those which we regard as of fixed value, and
call constants, we generally denote algebraically by letters from the beginning of the alphabet, such as a, b, or c; while those which we consider
as capable of growing, or (as mathematicians say) of “varying,” we denote by letters from the end of the alphabet, such as x, y, z, u, v, w,
or sometimes t.
Moreover, we are usually dealing with more than one variable at
once, and thinking of the way in which one variable depends on the
other: for instance, we think of the way in which the height reached
by a projectile depends on the time of attaining that height. Or we
are asked to consider a rectangle of given area, and to enquire how any
increase in the length of it will compel a corresponding decrease in the
breadth of it. Or we think of the way in which any variation in the
slope of a ladder will cause the height that it reaches, to vary.
Suppose we have got two such variables that depend one on the
other. An alteration in one will bring about an alteration in the other,
because of this dependence. Let us call one of the variables x, and the
10
CALCULUS MADE EASY
other that depends on it y.
Suppose we make x to vary, that is to say, we either alter it or
imagine it to be altered, by adding to it a bit which we call dx. We are
thus causing x to become x + dx. Then, because x has been altered,
y will have altered also, and will have become y + dy. Here the bit dy
may be in some cases positive, in others negative; and it won’t (except
by a miracle) be the same size as dx.
Take two examples.
(1) Let x and y be respectively the base and the height of a rightangled triangle (Fig. 4), of which the slope of the other side is fixed
dy
y
y
30◦
x
dx
Fig. 4.
at 30◦ . If we suppose this triangle to expand and yet keep its angles
the same as at first, then, when the base grows so as to become x + dx,
the height becomes y + dy. Here, increasing x results in an increase
of y. The little triangle, the height of which is dy, and the base of which
is dx, is similar to the original triangle; and it is obvious that the value
dy
y
of the ratio
is the same as that of the ratio . As the angle is 30◦
dx
x
it will be seen that here
dy
1
=
.
dx
1.73
ON RELATIVE GROWINGS
11
(2) Let x represent, in Fig. 5, the horizontal distance, from a wall,
of the bottom end of a ladder, AB, of fixed length; and let y be the
B
y
O
x
A
Fig. 5.
height it reaches up the wall. Now y clearly depends on x. It is easy to
see that, if we pull the bottom end A a bit further from the wall, the
top end B will come down a little lower. Let us state this in scientific
language. If we increase x to x + dx, then y will become y − dy; that is,
when x receives a positive increment, the increment which results to y
is negative.
Yes, but how much? Suppose the ladder was so long that when the
bottom end A was 19 inches from the wall the top end B reached just
15 feet from the ground. Now, if you were to pull the bottom end out
1 inch more, how much would the top end come down? Put it all into
inches: x = 19 inches, y = 180 inches. Now the increment of x which
we call dx, is 1 inch: or x + dx = 20 inches.
CALCULUS MADE EASY
12
How much will y be diminished? The new height will be y − dy. If
we work out the height by Euclid I. 47, then we shall be able to find
how much dy will be. The length of the ladder is
p
(180)2 + (19)2 = 181 inches.
Clearly then, the new height, which is y − dy, will be such that
(y − dy)2 = (181)2 − (20)2 = 32761 − 400 = 32361,
√
y − dy = 32361 = 179.89 inches.
Now y is 180, so that dy is 180 − 179.89 = 0.11 inch.
So we see that making dx an increase of 1 inch has resulted in
making dy a decrease of 0.11 inch.
And the ratio of dy to dx may be stated thus:
0.11
dy
=−
.
dx
1
It is also easy to see that (except in one particular position) dy will
be of a different size from dx.
Now right through the differential calculus we are hunting, hunting,
hunting for a curious thing, a mere ratio, namely, the proportion which
dy bears to dx when both of them are indefinitely small.
dy
when
dx
y and x are related to each other in some way, so that whenever x varies
It should be noted here that we can only find this ratio
y does vary also. For instance, in the first example just taken, if the
base x of the triangle be made longer, the height y of the triangle
becomes greater also, and in the second example, if the distance x of
the foot of the ladder from the wall be made to increase, the height y
ON RELATIVE GROWINGS
13
reached by the ladder decreases in a corresponding manner, slowly at
first, but more and more rapidly as x becomes greater. In these cases
the relation between x and y is perfectly definite, it can be expressed
y
= tan 30◦ and x2 + y 2 = l2 (where l is the
mathematically, being
x
dy
length of the ladder) respectively, and
has the meaning we found in
dx
each case.
If, while x is, as before, the distance of the foot of the ladder from
the wall, y is, instead of the height reached, the horizontal length of
the wall, or the number of bricks in it, or the number of years since it
was built, any change in x would naturally cause no change whatever
dy
has no meaning whatever, and it is not possible
in y; in this case
dx
to find an expression for it. Whenever we use differentials dx, dy,
dz, etc., the existence of some kind of relation between x, y, z, etc., is
implied, and this relation is called a “function” in x, y, z, etc.; the two
y
expressions given above, for instance, namely = tan 30◦ and x2 +y 2 =
x
l2 , are functions of x and y. Such expressions contain implicitly (that
is, contain without distinctly showing it) the means of expressing either
x in terms of y or y in terms of x, and for this reason they are called
implicit functions in x and y; they can be respectively put into the
forms
and
y
y = x tan 30◦ or x =
tan 30◦
p
√
y = l2 − x2 or x = l2 − y 2 .
These last expressions state explicitly (that is, distinctly) the value
of x in terms of y, or of y in terms of x, and they are for this reason
called explicit functions of x or y. For example x2 + 3 = 2y − 7 is an
CALCULUS MADE EASY
14
x2 + 10
(explicit
implicit function in x and y; it may be written y =
2
√
function of x) or x = 2y − 10 (explicit function of y). We see that
an explicit function in x, y, z, etc., is simply something the value of
which changes when x, y, z, etc., are changing, either one at the time
or several together. Because of this, the value of the explicit function
is called the dependent variable, as it depends on the value of the other
variable quantities in the function; these other variables are called the
independent variables because their value is not determined from the
value assumed by the function. For example, if u = x2 sin θ, x and θ
are the independent variables, and u is the dependent variable.
Sometimes the exact relation between several quantities x, y, z either is not known or it is not convenient to state it; it is only known,
or convenient to state, that there is some sort of relation between these
variables, so that one cannot alter either x or y or z singly without
affecting the other quantities; the existence of a function in x, y, z
is then indicated by the notation F (x, y, z) (implicit function) or by
x = F (y, z), y = F (x, z) or z = F (x, y) (explicit function). Sometimes
the letter f or φ is used instead of F , so that y = F (x), y = f (x) and
y = φ(x) all mean the same thing, namely, that the value of y depends
on the value of x in some way which is not stated.
dy
“the differential coefficient of y with respect
We call the ratio
dx
to x.” It is a solemn scientific name for this very simple thing. But
we are not going to be frightened by solemn names, when the things
themselves are so easy. Instead of being frightened we will simply pronounce a brief curse on the stupidity of giving long crack-jaw names;
and, having relieved our minds, will go on to the simple thing itself,
ON RELATIVE GROWINGS
15
dy
.
dx
In ordinary algebra which you learned at school, you were always
namely the ratio
hunting after some unknown quantity which you called x or y; or sometimes there were two unknown quantities to be hunted for simultaneously. You have now to learn to go hunting in a new way; the fox being
now neither x nor y. Instead of this you have to hunt for this curious
dy
dy
. The process of finding the value of
is called “difcub called
dx
dx
ferentiating.” But, remember, what is wanted is the value of this ratio
when both dy and dx are themselves indefinitely small. The true value
of the differential coefficient is that to which it approximates in the
limiting case when each of them is considered as infinitesimally minute.
dy
.
Let us now learn how to go in quest of
dx
CALCULUS MADE EASY
16
NOTE TO CHAPTER III.
How to read Differentials.
It will never do to fall into the schoolboy error of thinking that dx
means d times x, for d is not a factor—it means “an element of” or “a
bit of” whatever follows. One reads dx thus: “dee-eks.”
In case the reader has no one to guide him in such matters it may
here be simply said that one reads differential coefficients in the following way. The differential coefficient
dy
is read “dee-wy by dee-eks,” or “dee-wy over dee-eks.”
dx
du
is read “dee-you by dee-tee.”
So also
dt
Second differential coefficients will be met with later on. They are
like this:
d2 y
; which is read “dee-two-wy over dee-eks-squared,”
dx2
and it means that the operation of differentiating y with respect to x
has been (or has to be) performed twice over.
Another way of indicating that a function has been differentiated is
by putting an accent to the symbol of the function. Thus if y = F (x),
which means that y is some unspecified
function of x (see p. 13), we may
d F (x)
write F 0 (x) instead of
. Similarly, F 00 (x) will mean that the
dx
original function F (x) has been differentiated twice over with respect
to x.
CHAPTER IV.
SIMPLEST CASES.
Now let us see how, on first principles, we can differentiate some simple
algebraical expression.
Case 1.
Let us begin with the simple expression y = x2 . Now remember
that the fundamental notion about the calculus is the idea of growing.
Mathematicians call it varying. Now as y and x2 are equal to one
another, it is clear that if x grows, x2 will also grow. And if x2 grows,
then y will also grow. What we have got to find out is the proportion
between the growing of y and the growing of x. In other words our task
is to find out the ratio between dy and dx, or, in brief, to find the value
dy
.
of
dx
Let x, then, grow a little bit bigger and become x + dx; similarly,
y will grow a bit bigger and will become y+dy. Then, clearly, it will still
be true that the enlarged y will be equal to the square of the enlarged x.
Writing this down, we have:
y + dy = (x + dx)2 .
CALCULUS MADE EASY
18
Doing the squaring we get:
y + dy = x2 + 2x · dx + (dx)2 .
What does (dx)2 mean? Remember that dx meant a bit—a little
bit—of x. Then (dx)2 will mean a little bit of a little bit of x; that
is, as explained above (p. 4), it is a small quantity of the second order
of smallness. It may therefore be discarded as quite inconsiderable in
comparison with the other terms. Leaving it out, we then have:
y + dy = x2 + 2x · dx.
Now y = x2 ; so let us subtract this from the equation and we have
left
dy = 2x · dx.
Dividing across by dx, we find
dy
= 2x.
dx
Now this ∗ is what we set out to find. The ratio of the growing of y
to the growing of x is, in the case before us, found to be 2x.
dy
is the result of differentiating y with respect to x. Difdx
ferentiating means finding the differential coefficient. Suppose we had some other
∗
N.B.—This ratio
function of x, as, for example, u = 7x2 + 3. Then if we were told to differentiate this
du
d(7x2 + 3)
with respect to x, we should have to find
, or, what is the same thing,
.
dx
dx
On the other hand, we may have a case in which time was the independent variable
(see p. 14), such as this: y = b + 21 at2 . Then, if we were told to differentiate it, that
means we must find its differential coefficient with respect to t. So that then our
d(b + 12 at2 )
dy
business would be to try to find
, that is, to find
.
dt
dt
SIMPLEST CASES
19
Numerical example.
Suppose x = 100 and ∴ y = 10, 000. Then let x grow till it becomes
101 (that is, let dx = 1). Then the enlarged y will be 101 × 101 =
10, 201. But if we agree that we may ignore small quantities of the
second order, 1 may be rejected as compared with 10, 000; so we may
round off the enlarged y to 10, 200. y has grown from 10, 000 to 10, 200;
the bit added on is dy, which is therefore 200.
200
dy
=
= 200. According to the algebra-working of the previous
dx
1
dy
paragraph, we find
= 2x. And so it is; for x = 100 and 2x = 200.
dx
But, you will say, we neglected a whole unit.
Well, try again, making dx a still smaller bit.
Try dx =
1
.
10
Then x + dx = 100.1, and
(x + dx)2 = 100.1 × 100.1 = 10, 020.01.
Now the last figure 1 is only one-millionth part of the 10, 000, and
is utterly negligible; so we may take 10, 020 without the little decimal
dy
20
at the end. And this makes dy = 20; and
=
= 200, which is
dx
0.1
still the same as 2x.
Case 2.
Try differentiating y = x3 in the same way.
We let y grow to y + dy, while x grows to x + dx.
Then we have
y + dy = (x + dx)3 .
Doing the cubing we obtain
y + dy = x3 + 3x2 · dx + 3x(dx)2 + (dx)3 .
CALCULUS MADE EASY
20
Now we know that we may neglect small quantities of the second
and third orders; since, when dy and dx are both made indefinitely
small, (dx)2 and (dx)3 will become indefinitely smaller by comparison.
So, regarding them as negligible, we have left:
y + dy = x3 + 3x2 · dx.
But y = x3 ; and, subtracting this, we have:
dy = 3x2 · dx,
dy
= 3x2 .
dx
and
Case 3.
Try differentiating y = x4 . Starting as before by letting both y and x
grow a bit, we have:
y + dy = (x + dx)4 .
Working out the raising to the fourth power, we get
y + dy = x4 + 4x3 dx + 6x2 (dx)2 + 4x(dx)3 + (dx)4 .
Then striking out the terms containing all the higher powers of dx,
as being negligible by comparison, we have
y + dy = x4 + 4x3 dx.
Subtracting the original y = x4 , we have left
dy = 4x3 dx,
and
dy
= 4x3 .
dx
SIMPLEST CASES
21
Now all these cases are quite easy. Let us collect the results to see if
we can infer any general rule. Put them in two columns, the values of y
dy
in the other: thus
in one and the corresponding values found for
dx
y
dy
dx
x2
2x
x3
3x2
x4
4x3
Just look at these results: the operation of differentiating appears
to have had the effect of diminishing the power of x by 1 (for example
in the last case reducing x4 to x3 ), and at the same time multiplying by
a number (the same number in fact which originally appeared as the
power). Now, when you have once seen this, you might easily conjecture
how the others will run. You would expect that differentiating x5 would
give 5x4 , or differentiating x6 would give 6x5 . If you hesitate, try one
of these, and see whether the conjecture comes right.
Try y = x5 .
Then
y + dy = (x + dx)5
= x5 + 5x4 dx + 10x3 (dx)2 + 10x2 (dx)3
+ 5x(dx)4 + (dx)5 .
Neglecting all the terms containing small quantities of the higher
orders, we have left
y + dy = x5 + 5x4 dx,
CALCULUS MADE EASY
and subtracting
22
y = x5 leaves us
dy = 5x4 dx,
whence
dy
= 5x4 , exactly as we supposed.
dx
Following out logically our observation, we should conclude that if
we want to deal with any higher power,—call it n—we could tackle it
in the same way.
y = xn ,
Let
then, we should expect to find that
dy
= nx(n−1) .
dx
For example, let n = 8, then y = x8 ; and differentiating it would
dy
= 8x7 .
give
dx
And, indeed, the rule that differentiating xn gives as the result nxn−1
is true for all cases where n is a whole number and positive. [Expanding
(x + dx)n by the binomial theorem will at once show this.] But the
question whether it is true for cases where n has negative or fractional
values requires further consideration.
Case of a negative power.
Let y = x−2 . Then proceed as before:
y + dy = (x + dx)−2
−2
dx
−2
=x
1+
.
x
SIMPLEST CASES
23
Expanding this by the binomial theorem (see p. 137), we get
#
"
2
2(2
+
1)
dx
2
dx
− etc.
+
= x−2 1 −
x
1×2
x
= x−2 − 2x−3 · dx + 3x−4 (dx)2 − 4x−5 (dx)3 + etc.
So, neglecting the small quantities of higher orders of smallness, we
have:
y + dy = x−2 − 2x−3 · dx.
Subtracting the original y = x−2 , we find
dy = −2x−3 dx,
dy
= −2x−3 .
dx
And this is still in accordance with the rule inferred above.
Case of a fractional power.
1
Let y = x 2 . Then, as before,
1
dx 2
y + dy = (x + dx) = x 1 +
x
1
2
=
1
2
√
1 dx
1 (dx)2
√ + terms with higher
x+ √ −
2 x 8x x
powers of dx.
1
Subtracting the original y = x 2 , and neglecting higher powers we
have left:
dy =
1 dx
1 1
√ = x− 2 · dx,
2 x
2
24
CALCULUS MADE EASY
1 1
dy
= x− 2 . Agreeing with the general rule.
dx
2
Summary. Let us see how far we have got. We have arrived at the
and
following rule: To differentiate xn , multiply by the power and reduce
the power by one, so giving us nxn−1 as the result.
Exercises I.
(See p. 252 for Answers.)
Differentiate the following:
3
(1) y = x13
(2) y = x− 2
(3) y = x2a
(4) u = t2.4
(5) z =
√
3
u
r
(7) u =
(9) y =
5
1
x8
√
q 3
x
(6) y =
√
3 −5
x
(8) y = 2xa
r
(10) y =
n
1
xm
You have now learned how to differentiate powers of x. How easy it
is!
CHAPTER V.
NEXT STAGE. WHAT TO DO WITH CONSTANTS.
In our equations we have regarded x as growing, and as a result of x
being made to grow y also changed its value and grew. We usually
think of x as a quantity that we can vary; and, regarding the variation
of x as a sort of cause, we consider the resulting variation of y as an
effect. In other words, we regard the value of y as depending on that
of x. Both x and y are variables, but x is the one that we operate upon,
and y is the “dependent variable.” In all the preceding chapter we have
been trying to find out rules for the proportion which the dependent
variation in y bears to the variation independently made in x.
Our next step is to find out what effect on the process of differentiating is caused by the presence of constants, that is, of numbers which
don’t change when x or y change their values.
Added Constants.
Let us begin with some simple case of an added constant, thus:
Let
y = x3 + 5.
Just as before, let us suppose x to grow to x+dx and y to grow to y+dy.
CALCULUS MADE EASY
Then:
26
y + dy = (x + dx)3 + 5
= x3 + 3x2 dx + 3x(dx)2 + (dx)3 + 5.
Neglecting the small quantities of higher orders, this becomes
y + dy = x3 + 3x2 · dx + 5.
Subtract the original y = x3 + 5, and we have left:
dy = 3x2 dx.
dy
= 3x2 .
dx
So the 5 has quite disappeared. It added nothing to the growth
of x, and does not enter into the differential coefficient. If we had put 7,
or 700, or any other number, instead of 5, it would have disappeared.
So if we take the letter a, or b, or c to represent any constant, it will
simply disappear when we differentiate.
If the additional constant had been of negative value, such as
−5 or −b, it would equally have disappeared.
Multiplied Constants.
Take as a simple experiment this case:
Let y = 7x2 .
Then on proceeding as before we get:
y + dy = 7(x + dx)2
= 7{x2 + 2x · dx + (dx)2 }
= 7x2 + 14x · dx + 7(dx)2 .
27
WHAT TO DO WITH CONSTANTS
Then, subtracting the original y = 7x2 , and neglecting the last term,
we have
dy = 14x · dx.
dy
= 14x.
dx
Let us illustrate this example by working out the graphs of the
dy
equations y = 7x2 and
= 14x, by assigning to x a set of successive
dx
values, 0, 1, 2, 3, etc., and finding the corresponding values of y and
dy
.
of
dx
These values we tabulate as follows:
x
0
1
2
3
4
5
−1
−2
−3
y
0
7
28
63
112
175
7
28
63
dy
dx
0
14
28
42
56
70
−14
−28
−42
Now plot these values to some convenient scale, and we obtain the
two curves, Figs. 6 and 6a.
Carefully compare the two figures, and verify by inspection that
the height of the ordinate of the derived curve, Fig. 6a, is proportional
to the slope of the original curve,∗ Fig. 6, at the corresponding value
of x. To the left of the origin, where the original curve slopes negatively
(that is, downward from left to right) the corresponding ordinates of
the derived curve are negative.
Now if we look back at p. 18, we shall see that simply differentiating x2 gives us 2x. So that the differential coefficient of 7x2 is just
∗
See p. 76 about slopes of curves.
28
CALCULUS MADE EASY
dy
dx
y
200
150
150
100
100
50
50
−3 −2 −1
−3 −2 −1 0
1
2
3
4
5
x
0
1
2
3
4
5
−50
Fig. 6.—Graph of y = 7x2 .
Fig. 6a.—Graph of
dy
= 14x.
dx
7 times as big as that of x2 . If we had taken 8x2 , the differential coefficient would have come out eight times as great as that of x2 . If we put
y = ax2 , we shall get
dy
= a × 2x.
dx
dy
= a × nxn−1 .
dx
So that any mere multiplication by a constant reappears as a mere
If we had begun with y = axn , we should have had
multiplication when the thing is differentiated. And, what is true about
multiplication is equally true about division: for if, in the example
above, we had taken as the constant
the same
1
7
1
7
instead of 7, we should have had
come out in the result after differentiation.
Some Further Examples.
The following further examples, fully worked out, will enable you to
master completely the process of differentiation as applied to ordinary
x
WHAT TO DO WITH CONSTANTS
29
algebraical expressions, and enable you to work out by yourself the
examples given at the end of this chapter.
x5 3
− .
(1) Differentiate y =
7
5
3
is an added constant and vanishes (see p. 25).
5
We may then write at once
dy
1
= × 5 × x5−1 ,
dx
7
dy
5
= x4 .
dx
7
or
√
1√
a.
(2) Differentiate y = a x −
2
√
1√
The term
a vanishes, being an added constant; and as a x, in
2
1
the index form, is written ax 2 , we have
1
1
dy
1
a
= a × × x 2 −1 = × x− 2 ,
dx
2
2
dy
a
or
= √ .
dx
2 x
√
(3) If ay + bx = by − ax + (x + y) a2 − b2 ,
find the differential coefficient of y with respect to x.
As a rule an expression of this kind will need a little more knowledge
than we have acquired so far; it is, however, always worth while to try
whether the expression can be put in a simpler form.
First we must try to bring it into the form y = some expression
involving x only.
The expression may be written
√
(a − b)y + (a + b)x = (x + y) a2 − b2 .
30
CALCULUS MADE EASY
Squaring, we get
(a − b)2 y 2 + (a + b)2 x2 + 2(a + b)(a − b)xy = (x2 + y 2 + 2xy)(a2 − b2 ),
which simplifies to
(a − b)2 y 2 + (a + b)2 x2 = x2 (a2 − b2 ) + y 2 (a2 − b2 );
[(a − b)2 − (a2 − b2 )]y 2 = [(a2 − b2 ) − (a + b)2 ]x2 ,
or
2b(b − a)y 2 = −2b(b + a)x2 ;
that is
r
hence
y=
a+b
x and
a−b
dy
=
dx
r
a+b
.
a−b
(4) The volume of a cylinder of radius r and height h is given by
the formula V = πr2 h. Find the rate of variation of volume with the
radius when r = 5.5 in. and h = 20 in. If r = h, find the dimensions
of the cylinder so that a change of 1 in. in radius causes a change of
400 cub. in. in the volume.
The rate of variation of V with regard to r is
dV
= 2πrh.
dr
If r = 5.5 in. and h = 20 in. this becomes 690.8.
It means
that a change of radius of 1 inch will cause a change of volume of
690.8 cub. inch. This can be easily verified, for the volumes with
r = 5 and r = 6 are 1570 cub. in. and 2260.8 cub. in. respectively, and
2260.8 − 1570 = 690.8.
Also, if
r = h,
dV
= 2πr2 = 400 and r = h =
dr
r
400
= 7.98 in.
2π
WHAT TO DO WITH CONSTANTS
31
(5) The reading θ of a Féry’s Radiation pyrometer is related to the
Centigrade temperature t of the observed body by the relation
4
t
θ
=
,
θ1
t1
where θ1 is the reading corresponding to a known temperature t1 of the
observed body.
Compare the sensitiveness of the pyrometer at temperatures
800◦ C., 1000◦ C., 1200◦ C., given that it read 25 when the temperature
was 1000◦ C.
The sensitiveness is the rate of variation of the reading with the
dθ
. The formula may be written
temperature, that is
dt
θ=
and we have
25t4
θ1 4
t
=
,
t41
10004
dθ
100t3
t3
=
=
.
dt
10004
10, 000, 000, 000
When t = 800, 1000 and 1200, we get
respectively.
dθ
= 0.0512, 0.1 and 0.1728
dt
The sensitiveness is approximately doubled from 800◦ to 1000◦ , and
becomes three-quarters as great again up to 1200◦ .
Exercises II.
(See p. 252 for Answers.)
Differentiate the following:
(1) y = ax3 + 6.
3
(2) y = 13x 2 − c.
32
CALCULUS MADE EASY
1
1
1
(3) y = 12x 2 + c 2 .
(5) u =
1
(4) y = c 2 x 2 .
az n − 1
.
c
(6) y = 1.18t2 + 22.4.
Make up some other examples for yourself, and try your hand at
differentiating them.
(7) If lt and l0 be the lengths of a rod of iron at the temperatures
t◦ C. and 0◦ C. respectively, then lt = l0 (1+0.000012t). Find the change
of length of the rod per degree Centigrade.
(8) It has been found that if c be the candle power of an incandescent electric lamp, and V be the voltage, c = aV b , where a and b are
constants.
Find the rate of change of the candle power with the voltage, and
calculate the change of candle power per volt at 80, 100 and 120 volts
in the case of a lamp for which a = 0.5 × 10−10 and b = 6.
(9) The frequency n of vibration of a string of diameter D, length L
and specific gravity σ, stretched with a force T , is given by
r
gT
1
.
n=
DL πσ
Find the rate of change of the frequency when D, L, σ and T are
varied singly.
(10) The greatest external pressure P which a tube can support without collapsing is given by
P =
2E
1 − σ2
t3
,
D3
WHAT TO DO WITH CONSTANTS
33
where E and σ are constants, t is the thickness of the tube and D is
its diameter. (This formula assumes that 4t is small compared to D.)
Compare the rate at which P varies for a small change of thickness
and for a small change of diameter taking place separately.
(11) Find, from first principles, the rate at which the following vary
with respect to a change in radius:
(a) the circumference of a circle of radius r;
(b) the area of a circle of radius r;
(c) the lateral area of a cone of slant dimension l;
(d ) the volume of a cone of radius r and height h;
(e) the area of a sphere of radius r;
(f ) the volume of a sphere of radius r.
(12) The length L of an iron rod at the temperature T being given by
L = lt 1 + 0.000012(T − t) , where lt is the length at the temperature t,
find the rate of variation of the diameter D of an iron tyre suitable for
being shrunk on a wheel, when the temperature T varies.
CHAPTER VI.
SUMS, DIFFERENCES, PRODUCTS AND
QUOTIENTS.
We have learned how to differentiate simple algebraical functions such
as x2 + c or ax4 , and we have now to consider how to tackle the sum
of two or more functions.
For instance, let
y = (x2 + c) + (ax4 + b);
dy
be? How are we to go to work on this new job?
dx
The answer to this question is quite simple: just differentiate them,
what will its
one after the other, thus:
dy
= 2x + 4ax3 . (Ans.)
dx
If you have any doubt whether this is right, try a more general case,
working it by first principles. And this is the way.
Let y = u+v, where u is any function of x, and v any other function
of x. Then, letting x increase to x + dx, y will increase to y + dy; and
u will increase to u + du; and v to v + dv.
And we shall have:
y + dy = u + du + v + dv.
SUMS, DIFFERENCES, PRODUCTS
35
Subtracting the original y = u + v, we get
dy = du + dv,
and dividing through by dx, we get:
du dv
dy
=
+ .
dx
dx dx
This justifies the procedure. You differentiate each function separately and add the results. So if now we take the example of the
preceding paragraph, and put in the values of the two functions, we
shall have, using the notation shown (p. 16),
dy
d(x2 + c) d(ax4 + b)
=
+
dx
dx
dx
3
= 2x
+ 4ax ,
exactly as before.
If there were three functions of x, which we may call u, v and w, so
that
y = u + v + w;
then
du dv dw
dy
=
+
+
.
dx
dx dx dx
As for subtraction, it follows at once; for if the function v had itself
had a negative sign, its differential coefficient would also be negative;
so that by differentiating
y = u − v,
we should get
dy
du dv
=
− .
dx
dx dx
CALCULUS MADE EASY
36
But when we come to do with Products, the thing is not quite so
simple.
Suppose we were asked to differentiate the expression
y = (x2 + c) × (ax4 + b),
what are we to do? The result will certainly not be 2x × 4ax3 ; for it
is easy to see that neither c × ax4 , nor x2 × b, would have been taken
into that product.
Now there are two ways in which we may go to work.
First way. Do the multiplying first, and, having worked it out, then
differentiate.
Accordingly, we multiply together x2 + c and ax4 + b.
This gives ax6 + acx4 + bx2 + bc.
Now differentiate, and we get:
dy
= 6ax5 + 4acx3 + 2bx.
dx
Second way. Go back to first principles, and consider the equation
y = u × v;
where u is one function of x, and v is any other function of x. Then, if
x grows to be x + dx; and y to y + dy; and u becomes u + du, and v
becomes v + dv, we shall have:
y + dy = (u + du) × (v + dv)
= u · v + u · dv + v · du + du · dv.
SUMS, DIFFERENCES, PRODUCTS
37
Now du · dv is a small quantity of the second order of smallness, and
therefore in the limit may be discarded, leaving
y + dy = u · v + u · dv + v · du.
Then, subtracting the original y = u · v, we have left
dy = u · dv + v · du;
and, dividing through by dx, we get the result:
dv
du
dy
=u
+v .
dx
dx
dx
This shows that our instructions will be as follows: To differentiate
the product of two functions, multiply each function by the differential
coefficient of the other, and add together the two products so obtained.
You should note that this process amounts to the following: Treat u
as constant while you differentiate v; then treat v as constant while you
dy
differentiate u; and the whole differential coefficient
will be the sum
dx
of these two treatments.
Now, having found this rule, apply it to the concrete example which
was considered above.
We want to differentiate the product
(x2 + c) × (ax4 + b).
Call (x2 + c) = u; and (ax4 + b) = v.
CALCULUS MADE EASY
38
Then, by the general rule just established, we may write:
d(ax4 + b)
d(x2 + c)
dy
= (x2 + c)
+ (ax4 + b)
dx
dx
dx
= (x2 + c) 4ax3
+ (ax4 + b) 2x
= 4ax5 + 4acx3
+ 2ax5 + 2bx,
dy
= 6ax5 + 4acx3
dx
+ 2bx,
exactly as before.
Lastly, we have to differentiate quotients.
bx5 + c
Think of this example, y = 2
. In such a case it is no use to
x +a
try to work out the division beforehand, because x2 + a will not divide
into bx5 + c, neither have they any common factor. So there is nothing
for it but to go back to first principles, and find a rule.
So we will put
y=
u
;
v
where u and v are two different functions of the independent variable x.
Then, when x becomes x + dx, y will become y + dy; and u will become
u + du; and v will become v + dv. So then
y + dy =
u + du
.
v + dv
QUOTIENTS
39
Now perform the algebraic division, thus:
v + dv u + du
u · dv
u+
v
u du u · dv
+
−
v
v
v2
u · dv
v
du · dv
du +
v
du −
u · dv du · dv
−
v
v
u · dv u · dv · dv
−
−
v
v2
−
−
du · dv u · dv · dv
+
.
v
v2
As both these remainders are small quantities of the second order,
they may be neglected, and the division may stop here, since any further
remainders would be of still smaller magnitudes.
So we have got:
y + dy =
u du u · dv
;
+
−
v
v
v2
which may be written
=
u v · du − u · dv
+
.
v
v2
CALCULUS MADE EASY
Now subtract the original y =
40
u
, and we have left:
v
v · du − u · dv
;
v2
du
dv
v
−
u
dy
= dx 2 dx .
dx
v
dy =
whence
This gives us our instructions as to how to differentiate a quotient
of two functions. Multiply the divisor function by the differential coefficient of the dividend function; then multiply the dividend function by
the differential coefficient of the divisor function; and subtract. Lastly
divide by the square of the divisor function.
bx5 + c
,
Going back to our example y = 2
x +a
write
bx5 + c = u;
and
x2 + a = v.
Then
d(bx5 + c)
d(x2 + a)
5
−
(bx
+
c)
dy
dx
dx
=
2
2
dx
(x + a)
2
4
(x + a)(5bx ) − (bx5 + c)(2x)
=
,
(x2 + a)2
dy
3bx6 + 5abx4 − 2cx
=
. (Answer.)
dx
(x2 + a)2
(x2 + a)
The working out of quotients is often tedious, but there is nothing
difficult about it.
Some further examples fully worked out are given hereafter.
DIFFERENTIATION
(1) Differentiate y =
Being a constant,
a2
a 3 a2
x
+
x
−
.
b2
b
b2
a2
vanishes, and we have
b2
a
a2
dy
= 2 × 3 × x3−1 −
× 1 × x1−1 .
dx
b
b
But x1−1 = x0 = 1; so we get:
dy
3a 2 a2
= 2x − .
dx
b
b
√
√
√
3b 3 a
(2) Differentiate y = 2a bx3 −
− 2 ab.
x
Putting x in the index form, we get
√ 3
√
√
y = 2a bx 2 − 3b 3 ax−1 − 2 ab.
Now
or,
√
√
3
dy
= 2a b × 32 × x 2 −1 − 3b 3 a × (−1) × x−1−1 ;
dx
√
√
3b 3 a
dy
.
= 3a bx +
dx
x2
r
1
4.4
−√
− 27◦ .
5
2
θ
θ
2
1
This may be written: z = 1.8 θ− 3 − 4.4 θ− 5 − 27◦ .
(3) Differentiate z = 1.8 3
The 27◦ vanishes, and we have
or,
or,
1
2
dz
= 1.8 × − 32 × θ− 3 −1 − 4.4 × − 15 θ− 5 −1 ;
dθ
5
6
dz
= −1.2 θ− 3 + 0.88 θ− 5 ;
dθ
0.88
1.2
dz
= √
−√
.
5 6
3 5
dθ
θ
θ
41
CALCULUS MADE EASY
(4) Differentiate v = (3t2 − 1.2t + 1)3 .
42
A direct way of doing this will be explained later (see p. 66); but
we can nevertheless manage it now without any difficulty.
Developing the cube, we get
v = 27t6 − 32.4t5 + 39.96t4 − 23.328t3 − 13.32t2 − 3.6t + 1;
hence
dv
= 162t5 − 162t4 + 159.84t3 − 69.984t2 + 26.64t − 3.6.
dt
(5) Differentiate y = (2x − 3)(x + 1)2 .
d (x + 1)(x + 1)
dy
d(2x − 3)
= (2x − 3)
+ (x + 1)2
dx
dx
dx d(x + 1)
d(x + 1)
= (2x − 3) (x + 1)
+ (x + 1)
dx
dx
d(2x − 3)
+ (x + 1)2
dx
= 2(x + 1) (2x − 3) + (x + 1) = 2(x + 1)(3x − 2);
or, more simply, multiply out and then differentiate.
(6) Differentiate y = 0.5x3 (x − 3).
dy
d(x3 )
3 d(x − 3)
= 0.5 x
+ (x − 3)
dx
dx
dx
3
= 0.5 x + (x − 3) × 3x2 = 2x3 − 4.5x2 .
Same remarks as for preceding
example. √
1
1
(7) Differentiate w = θ +
θ+√ .
θ
θ
43
DIFFERENTIATION
This may be written
1
1
w = (θ + θ−1 )(θ 2 + θ− 2 ).
1
1
−2
−1
2
1
dw
)
− 12 d(θ + θ )
−1 d(θ + θ
2
= (θ + θ )
+ (θ + θ )
dθ
dθ
dθ
1
− 21
−1 1 − 21
1 − 32
2
= (θ + θ )( 2 θ − 2 θ ) + (θ + θ )(1 − θ−2 )
1
3
1
5
1
1
3
5
= 12 (θ 2 + θ− 2 − θ− 2 − θ− 2 ) + (θ 2 + θ− 2 − θ− 2 − θ− 2 )
√
1
1
1
3
1
=2
θ−√
+2 √ −√
.
θ
θ5
θ3
This, again, could be obtained more simply by multiplying the two
factors first, and differentiating afterwards. This is not, however, always
possible; see, for instance, p. 170, example 8, in which the rule for
differentiating a product must be used.
a
√
.
(8) Differentiate y =
1 + a x + a2 x
1
d(1 + ax 2 + a2 x)
(1
+
ax
+
a
x)
×
0
−
a
dy
dx
√
=
dx
(1 + a x + a2 x)2
1
2
2
1
=−
a( 12 ax− 2 + a2 )
1
(1 + ax 2 + a2 x)2
.
x2
.
(9) Differentiate y = 2
x +1
dy
(x2 + 1) 2x − x2 × 2x
2x
=
=
.
dx
(x2 + 1)2
(x2 + 1)2
√
a+ x
√ .
(10) Differentiate y =
a− x
44
CALCULUS MADE EASY
1
In the indexed form, y =
1
a + x2
.
1
a − x2
1
1
1
1
1
(a − x 2 )( 12 x− 2 ) − (a + x 2 )(− 12 x− 2 )
a − x2 + a + x2
dy
=
=
1 ;
1
1
dx
(a − x 2 )2
2(a − x 2 )2 x 2
a
dy
√ 2√ .
=
hence
dx
(a − x) x
√
3
1 − a t2
√
θ=
.
2
1 + a t3
2
1 − at 3
θ=
3 .
1 + at 2
(11) Differentiate
Now
3
1
2
1
(1 + at 2 )(− 23 at− 3 ) − (1 − at 3 ) × 32 at 2
dθ
=
3
dt
(1 + at 2 )2
√
√
4a
6
− 9a 2 t
5a2 t7 − √
3
t
√
.
=
2 3 2
6(1 + a t )
(12) A reservoir of square cross-section has sides sloping at an angle
of 45◦ with the vertical. The side of the bottom is 200 feet. Find an
expression for the quantity pouring in or out when the depth of water
varies by 1 foot; hence find, in gallons, the quantity withdrawn hourly
when the depth is reduced from 14 to 10 feet in 24 hours.
The volume of a frustum of pyramid of height H, and of bases A
√
H
and a, is V = (A + a + Aa). It is easily seen that, the slope being
3
45◦ , if the depth be h, the length of the side of the square surface of
the water is 200 + 2h feet, so that the volume of water is
h
4h3
[2002 + (200 + 2h)2 + 200(200 + 2h)] = 40, 000h + 400h2 +
.
3
3
DIFFERENTIATION
45
dV
= 40, 000 + 800h + 4h2 = cubic feet per foot of depth variation.
dh
dV
The mean level from 14 to 10 feet is 12 feet, when h = 12,
=
dh
50, 176 cubic feet.
Gallons per hour corresponding to a change of depth of 4 ft. in
4 × 50, 176 × 6.25
= 52, 267 gallons.
24 hours =
24
(13) The absolute pressure, in atmospheres, P , of saturated steam
5
40 + t
◦
at the temperature t C. is given by Dulong as being P =
140
as long as t is above 80◦ . Find the rate of variation of the pressure with
the temperature at 100◦ C.
Expand the numerator by the binomial theorem (see p. 137).
P =
hence
1
(405 + 5 × 404 t + 10 × 403 t2 + 10 × 402 t3 + 5 × 40t4 + t5 );
1405
1
dP
=
dt 537, 824 × 105
(5 × 404 + 20 × 403 t + 30 × 402 t2 + 20 × 40t3 + 5t4 ),
when t = 100 this becomes 0.036 atmosphere per degree Centigrade
change of temperature.
Exercises III.
(See the Answers on p. 253.)
(1) Differentiate
x2
x3
+
+ ···.
1×2 1×2×3
(b) y = ax2 + bx + c.
(c) y = (x + a)2 .
(a) u = 1 + x +
(d ) y = (x + a)3 .
CALCULUS MADE EASY
46
dw
.
dt
(3) Find the differential coefficient of
(2) If w = at − 21 bt2 , find
y = (x +
√
−1) × (x −
√
−1).
(4) Differentiate
y = (197x − 34x2 ) × (7 + 22x − 83x3 ).
dx
.
dy
(6) Differentiate y = 1.3709x × (112.6 + 45.202x2 ).
(5) If x = (y + 3) × (y + 5), find
Find the differential coefficients of
(7) y =
2x + 3
.
3x + 2
(8) y =
1 + x + 2x2 + 3x3
.
1 + x + 2x2
(9) y =
ax + b
.
cx + d
(10) y =
xn + a
.
x−n + b
(11) The temperature t of the filament of an incandescent electric
lamp is connected to the current passing through the lamp by the relation
C = a + bt + ct2 .
Find an expression giving the variation of the current corresponding
to a variation of temperature.
(12) The following formulae have been proposed to express the relation between the electric resistance R of a wire at the temperature
DIFFERENTIATION
47
t◦ C., and the resistance R0 of that same wire at 0◦ Centigrade, a, b, c
being constants.
R = R0 (1 + at + bt2 ).
√
R = R0 (1 + at + b t).
R = R0 (1 + at + bt2 )−1 .
Find the rate of variation of the resistance with regard to temperature as given by each of these formulae.
(13) The electromotive-force E of a certain type of standard cell has
been found to vary with the temperature t according to the relation
E = 1.4340 1 − 0.000814(t − 15) + 0.000007(t − 15)2 volts.
Find the change of electromotive-force per degree, at 15◦ , 20◦
and 25◦ .
(14) The electromotive-force necessary to maintain an electric arc of
length l with a current of intensity i has been found by Mrs. Ayrton to
be
E = a + bl +
c + kl
,
i
where a, b, c, k are constants.
Find an expression for the variation of the electromotive-force
(a) with regard to the length of the arc; (b) with regard to the strength
of the current.
CHAPTER VII.
SUCCESSIVE DIFFERENTIATION.
Let us try the effect of repeating several times over the operation of
differentiating a function (see p. 13). Begin with a concrete case.
Let y = x5 .
First differentiation,
5x4 .
Second differentiation, 5 × 4x3
= 20x3 .
Third differentiation, 5 × 4 × 3x2
= 60x2 .
Fourth differentiation, 5 × 4 × 3 × 2x
= 120x.
Fifth differentiation,
Sixth differentiation,
5 × 4 × 3 × 2 × 1 = 120.
= 0.
There is a certain notation, with which we are already acquainted
(see p. 14), used by some writers, that is very convenient. This is
to employ the general symbol f (x) for any function of x. Here the
symbol f ( ) is read as “function of,” without saying what particular
function is meant. So the statement y = f (x) merely tells us that y is
a function of x, it may be x2 or axn , or cos x or any other complicated
function of x.
SUCCESSIVE DIFFERENTIATION
49
The corresponding symbol for the differential coefficient is f 0 (x),
dy
. This is called the “derived function”
which is simpler to write than
dx
of x.
Suppose we differentiate over again, we shall get the “second derived
function” or second differential coefficient, which is denoted by f 00 (x);
and so on.
Now let us generalize.
Let y = f (x) = xn .
First differentiation,
f 0 (x) = nxn−1 .
Second differentiation,
f 00 (x) = n(n − 1)xn−2 .
Third differentiation,
Fourth differentiation,
f 000 (x) = n(n − 1)(n − 2)xn−3 .
f 0000 (x) = n(n − 1)(n − 2)(n − 3)xn−4 .
etc., etc.
But this is not the only way of indicating successive differentiations.
For,
if the original function be
y = f (x);
dy
= f 0 (x);
dx
dy
d
dx
twice differentiating gives
= f 00 (x);
dx
d2 y
d2 y
and this is more conveniently written as
,
or
more
usually
.
(dx)2
dx2
d3 y
Similarly, we may write as the result of thrice differentiating,
=
dx3
000
f (x).
once differentiating gives
CALCULUS MADE EASY
Examples.
Now let us try y = f (x) = 7x4 + 3.5x3 − 12 x2 + x − 2.
dy
dx
d2 y
dx2
d3 y
dx3
d4 y
dx4
d5 y
dx5
= f 0 (x) = 28x3 + 10.5x2 − x + 1,
= f 00 (x) = 84x2 + 21x − 1,
= f 000 (x) = 168x + 21,
= f 0000 (x) = 168,
= f 00000 (x) = 0.
In a similar manner if y = φ(x) = 3x(x2 − 4),
dy
= 3 x × 2x + (x2 − 4) × 1 = 3(3x2 − 4),
dx
d2 y
φ00 (x) = 2 = 3 × 6x = 18x,
dx
d3 y
φ000 (x) = 3 = 18,
dx
d4 y
φ0000 (x) = 4 = 0.
dx
φ0 (x) =
Exercises IV. (See page 253 for Answers.)
dy
d2 y
Find
and 2 for the following expressions:
dx
dx
(1) y = 17x + 12x2 .
(3) y = 1 +
(2) y =
x2 + a
.
x+a
x
x2
x3
x4
+
+
+
.
1 1×2 1×2×3 1×2×3×4
50
SUCCESSIVE DIFFERENTIATION
51
(4) Find the 2nd and 3rd derived functions in the Exercises III.
(p. 45), No. 1 to No. 7, and in the Examples given (p. 40), No. 1 to
No. 7.
CHAPTER VIII.
WHEN TIME VARIES.
Some of the most important problems of the calculus are those where
time is the independent variable, and we have to think about the values
of some other quantity that varies when the time varies. Some things
grow larger as time goes on; some other things grow smaller. The distance that a train has got from its starting place goes on ever increasing
as time goes on. Trees grow taller as the years go by. Which is growing
at the greater rate; a plant 12 inches high which in one month becomes
14 inches high, or a tree 12 feet high which in a year becomes 14 feet
high?
In this chapter we are going to make much use of the word rate.
Nothing to do with poor-rate, or water-rate (except that even here the
word suggests a proportion—a ratio—so many pence in the pound).
Nothing to do even with birth-rate or death-rate, though these words
suggest so many births or deaths per thousand of the population. When
a motor-car whizzes by us, we say: What a terrific rate! When a
spendthrift is flinging about his money, we remark that that young
man is living at a prodigious rate. What do we mean by rate? In
both these cases we are making a mental comparison of something that
is happening, and the length of time that it takes to happen. If the
WHEN TIME VARIES
53
motor-car flies past us going 10 yards per second, a simple bit of mental
arithmetic will show us that this is equivalent—while it lasts—to a rate
of 600 yards per minute, or over 20 miles per hour.
Now in what sense is it true that a speed of 10 yards per second
is the same as 600 yards per minute? Ten yards is not the same as
600 yards, nor is one second the same thing as one minute. What we
mean by saying that the rate is the same, is this: that the proportion
borne between distance passed over and time taken to pass over it, is
the same in both cases.
Take another example. A man may have only a few pounds in his
possession, and yet be able to spend money at the rate of millions
a year—provided he goes on spending money at that rate for a few
minutes only. Suppose you hand a shilling over the counter to pay
for some goods; and suppose the operation lasts exactly one second.
Then, during that brief operation, you are parting with your money
at the rate of 1 shilling per second, which is the same rate as £3 per
minute, or £180 per hour, or £4320 per day, or £1, 576, 800 per year!
If you have £10 in your pocket, you can go on spending money at the
rate of a million a year for just 5 14 minutes.
It is said that Sandy had not been in London above five minutes
when “bang went saxpence.” If he were to spend money at that rate
all day long, say for 12 hours, he would be spending 6 shillings an hour,
or £3. 12s. per day, or £21. 12s. a week, not counting the Sawbbath.
Now try to put some of these ideas into differential notation.
Let y in this case stand for money, and let t stand for time.
If you are spending money, and the amount you spend in a short
54
CALCULUS MADE EASY
dy
time dt be called dy, the rate of spending it will be , or rather, should
dt
dy
be written with a minus sign, as − , because dy is a decrement, not an
dt
increment. But money is not a good example for the calculus, because
it generally comes and goes by jumps, not by a continuous flow—you
may earn £200 a year, but it does not keep running in all day long
in a thin stream; it comes in only weekly, or monthly, or quarterly, in
lumps: and your expenditure also goes out in sudden payments.
A more apt illustration of the idea of a rate is furnished by the
speed of a moving body. From London (Euston station) to Liverpool is
200 miles. If a train leaves London at 7 o’clock, and reaches Liverpool
at 11 o’clock, you know that, since it has travelled 200 miles in 4 hours,
its average rate must have been 50 miles per hour; because
200
4
=
50
.
1
Here you are really making a mental comparison between the distance
passed over and the time taken to pass over it. You are dividing one
by the other. If y is the whole distance, and t the whole time, clearly
y
the average rate is . Now the speed was not actually constant all the
t
way: at starting, and during the slowing up at the end of the journey,
the speed was less. Probably at some part, when running downhill,
the speed was over 60 miles an hour. If, during any particular element
of time dt, the corresponding element of distance passed over was dy,
dy
then at that part of the journey the speed was
. The rate at which
dt
one quantity (in the present instance, distance) is changing in relation
to the other quantity (in this case, time) is properly expressed, then,
by stating the differential coefficient of one with respect to the other.
A velocity, scientifically expressed, is the rate at which a very small
distance in any given direction is being passed over; and may therefore
WHEN TIME VARIES
55
be written
dy
.
dt
But if the velocity v is not uniform, then it must be either increasing
v=
or else decreasing. The rate at which a velocity is increasing is called the
acceleration. If a moving body is, at any particular instant, gaining an
additional velocity dv in an element of time dt, then the acceleration a
at that instant may be written
a=
but dv is itself d
dv
;
dt
dy
. Hence we may put
dt
dy
d
dt
a=
;
dt
d2 y
;
dt2
or the acceleration is the second differential coefficient of the distance,
and this is usually written a =
with respect to time. Acceleration is expressed as a change of velocity
in unit time, for instance, as being so many feet per second per second;
the notation used being feet ÷ second2 .
When a railway train has just begun to move, its velocity v is small;
but it is rapidly gaining speed—it is being hurried up, or accelerated,
d2 y
by the effort of the engine. So its 2 is large. When it has got up its
dt
d2 y
top speed it is no longer being accelerated, so that then 2 has fallen
dt
to zero. But when it nears its stopping place its speed begins to slow
down; may, indeed, slow down very quickly if the brakes are put on,
CALCULUS MADE EASY
56
and during this period of deceleration or slackening of pace, the value
d2 y
dv
, that is, of 2 will be negative.
of
dt
dt
To accelerate a mass m requires the continuous application of force.
The force necessary to accelerate a mass is proportional to the mass,
and it is also proportional to the acceleration which is being imparted.
Hence we may write for the force f , the expression
f = ma;
dv
;
dt
d2 y
f =m 2.
dt
or
f =m
or
The product of a mass by the speed at which it is going is called its
momentum, and is in symbols mv. If we differentiate momentum with
d(mv)
for the rate of change of momentum.
respect to time we shall get
dt
dv
But, since m is a constant quantity, this may be written m , which
dt
we see above is the same as f . That is to say, force may be expressed
either as mass times acceleration, or as rate of change of momentum.
Again, if a force is employed to move something (against an equal
and opposite counter-force), it does work ; and the amount of work done
is measured by the product of the force into the distance (in its own
direction) through which its point of application moves forward. So if
a force f moves forward through a length y, the work done (which we
may call w) will be
w = f × y;
where we take f as a constant force. If the force varies at different
parts of the range y, then we must find an expression for its value from
WHEN TIME VARIES
57
point to point. If f be the force along the small element of length dy,
the amount of work done will be f × dy. But as dy is only an element
of length, only an element of work will be done. If we write w for work,
then an element of work will be dw; and we have
dw = f × dy;
which may be written
dw = ma · dy;
or
or
d2 y
· dy;
dt2
dv
dw = m · dy.
dt
dw = m
Further, we may transpose the expression and write
dw
= f.
dy
This gives us yet a third definition of force; that if it is being used
to produce a displacement in any direction, the force (in that direction)
is equal to the rate at which work is being done per unit of length in
that direction. In this last sentence the word rate is clearly not used in
its time-sense, but in its meaning as ratio or proportion.
Sir Isaac Newton, who was (along with Leibnitz) an inventor of the
methods of the calculus, regarded all quantities that were varying as
flowing; and the ratio which we nowadays call the differential coefficient
he regarded as the rate of flowing, or the fluxion of the quantity in question. He did not use the notation of the dy and dx, and dt (this was due
to Leibnitz), but had instead a notation of his own. If y was a quantity
CALCULUS MADE EASY
58
that varied, or “flowed,” then his symbol for its rate of variation (or
“fluxion”) was ẏ. If x was the variable, then its fluxion was called ẋ.
The dot over the letter indicated that it had been differentiated. But
this notation does not tell us what is the independent variable with
dy
respect to which the differentiation has been effected. When we see
dt
dy
we know that y is to be differentiated with respect to t. If we see
dx
we know that y is to be differentiated with respect to x. But if we see
merely ẏ, we cannot tell without looking at the context whether this is
dy
dy
dy
or
or
, or what is the other variable. So, therefore,
to mean
dx
dt
dz
this fluxional notation is less informing than the differential notation,
and has in consequence largely dropped out of use. But its simplicity
gives it an advantage if only we will agree to use it for those cases exclusively where time is the independent variable. In that case ẏ will
du
d2 x
dy
and u̇ will mean
; and ẍ will mean 2 .
mean
dt
dt
dt
Adopting this fluxional notation we may write the mechanical equations considered in the paragraphs above, as follows:
distance
x,
velocity
v = ẋ,
acceleration
a = v̇ = ẍ,
force
f = mv̇ = mẍ,
work
w = x × mẍ.
Examples.
(1) A body moves so that the distance x (in feet), which it travels
from a certain point O, is given by the relation x = 0.2t2 + 10.4, where
t is the time in seconds elapsed since a certain instant. Find the velocity
WHEN TIME VARIES
59
and acceleration 5 seconds after the body began to move, and also
find the corresponding values when the distance covered is 100 feet.
Find also the average velocity during the first 10 seconds of its motion.
(Suppose distances and motion to the right to be positive.)
x = 0.2t2 + 10.4
Now
v = ẋ =
dx
= 0.4t;
dt
and a = ẍ =
d2 x
= 0.4 = constant.
dt2
When t = 0, x = 10.4 and v = 0. The body started from a point
10.4 feet to the right of the point O; and the time was reckoned from
the instant the body started.
When t = 5, v = 0.4 × 5 = 2 ft./sec.; a = 0.4 ft./sec2 .
When x = 100, 100 = 0.2t2 + 10.4, or t2 = 448, and t = 21.17 sec.;
v = 0.4 × 21.17 = 8.468 ft./sec.
When t = 10,
distance travelled = 0.2 × 102 + 10.4 − 10.4 = 20 ft.
Average velocity =
20
10
= 2 ft./sec.
(It is the same velocity as the velocity at the middle of the interval,
t = 5; for, the acceleration being constant, the velocity has varied
uniformly from zero when t = 0 to 4 ft./sec. when t = 10.)
(2) In the above problem let us suppose
x = 0.2t2 + 3t + 10.4.
dx
v = ẋ =
= 0.4t + 3;
dt
d2 x
a = ẍ = 2 = 0.4 = constant.
dt
When t = 0, x = 10.4 and v = 3 ft./sec, the time is reckoned from
the instant at which the body passed a point 10.4 ft. from the point O,
CALCULUS MADE EASY
60
its velocity being then already 3 ft./sec. To find the time elapsed since
it began moving, let v = 0; then 0.4t + 3 = 0, t = − .43 = −7.5 sec.
The body began moving 7.5 sec. before time was begun to be observed;
5 seconds after this gives t = −2.5 and v = 0.4 × −2.5 + 3 = 2 ft./sec.
When x = 100 ft.,
100 = 0.2t2 + 3t + 10.4;
or t2 + 15t − 448 = 0;
hence t = 14.95 sec., v = 0.4 × 14.95 + 3 = 8.98 ft./sec.
To find the distance travelled during the 10 first seconds of the
motion one must know how far the body was from the point O when it
started.
When t = −7.5,
x = 0.2 × (−7.5)2 − 3 × 7.5 + 10.4 = −0.85 ft.,
that is 0.85 ft. to the left of the point O.
Now, when t = 2.5,
x = 0.2 × 2.52 + 3 × 2.5 + 10.4 = 19.15.
So, in 10 seconds, the distance travelled was 19.15 + 0.85 = 20 ft.,
and
the average velocity =
20
10
= 2 ft./sec.
(3) Consider a similar problem when the distance is given by x =
0.2t2 − 3t + 10.4. Then v = 0.4t − 3, a = 0.4 = constant. When
t = 0, x = 10.4 as before, and v = −3; so that the body was moving
in the direction opposite to its motion in the previous cases. As the
acceleration is positive, however, we see that this velocity will decrease
WHEN TIME VARIES
61
as time goes on, until it becomes zero, when v = 0 or 0.4t − 3 = 0; or
t = 7.5 sec. After this, the velocity becomes positive; and 5 seconds
after the body started, t = 12.5, and
v = 0.4 × 12.5 − 3 = 2 ft./sec.
When x = 100,
100 = 0.2t2 − 3t + 10.4,
or t2 − 15t − 448 = 0,
t = 29.95; v = 0.4 × 29.95 − 3 = 8.98 ft./sec.
and
When v is zero, x = 0.2 × 7.52 − 3 × 7.5 + 10.4 = −0.85, informing
us that the body moves back to 0.85 ft. beyond the point O before it
stops. Ten seconds later
t = 17.5 and x = 0.2 × 17.52 − 3 × 17.5 + 10.4 = 19.15.
The distance travelled = .85 + 19.15 = 20.0, and the average velocity
is again 2 ft./sec.
(4) Consider yet another problem of the same sort with x = 0.2t3 −
3t2 + 10.4; v = 0.6t2 − 6t; a = 1.2t − 6. The acceleration is no more
constant.
When t = 0, x = 10.4, v = 0, a = −6. The body is at rest, but just
ready to move with a negative acceleration, that is to gain a velocity
towards the point O.
(5) If we have x = 0.2t3 − 3t + 10.4, then v = 0.6t2 − 3, and a = 1.2t.
When t = 0, x = 10.4; v = −3; a = 0.
The body is moving towards the point O with a velocity of 3 ft./sec.,
and just at that instant the velocity is uniform.
CALCULUS MADE EASY
62
We see that the conditions of the motion can always be at once
ascertained from the time-distance equation and its first and second
derived functions. In the last two cases the mean velocity during the
first 10 seconds and the velocity 5 seconds after the start will no more
be the same, because the velocity is not increasing uniformly, the acceleration being no longer constant.
(6) The angle θ (in radians) turned through by a wheel is given by
θ = 3 + 2t − 0.1t3 , where t is the time in seconds from a certain instant;
find the angular velocity ω and the angular acceleration α, (a) after
1 second; (b) after it has performed one revolution. At what time is it
at rest, and how many revolutions has it performed up to that instant?
Writing for the acceleration
ω = θ̇ =
dθ
= 2 − 0.3t2 ,
dt
α = θ̈ =
d2 θ
= −0.6t.
dt2
When t = 0, θ = 3; ω = 2 rad./sec.; α = 0.
When t = 1,
ω = 2 − 0.3 = 1.7 rad./sec.;
α = −0.6 rad./sec2 .
This is a retardation; the wheel is slowing down.
After 1 revolution
θ = 2π = 6.28;
6.28 = 3 + 2t − 0.1t3 .
By plotting the graph, θ = 3 + 2t − 0.1t3 , we can get the value or
values of t for which θ = 6.28; these are 2.11 and 3.03 (there is a third
negative value).
WHEN TIME VARIES
63
When t = 2.11,
ω = 2 − 1.34 = 0.66 rad./sec.;
θ = 6.28;
α = −1.27 rad./sec2 .
When t = 3.03,
θ = 6.28;
ω = 2 − 2.754 = −0.754 rad./sec.;
α = −1.82 rad./sec2 .
The velocity is reversed. The wheel is evidently at rest between
these two instants; it is at rest when ω = 0, that is when 0 = 2 − 0.3t3 ,
or when t = 2.58 sec., it has performed
3 + 2 × 2.58 − 0.1 × 2.583
θ
=
= 1.025 revolutions.
2π
6.28
Exercises V.
(See page 255 for Answers.)
(1) If y = a + bt2 + ct4 ; find
Ans.
dy
d2 y
and 2 .
dt
dt
dy
= 2bt + 4ct3 ;
dt
d2 y
= 2b + 12ct2 .
dt2
(2) A body falling freely in space describes in t seconds a space s,
in feet, expressed by the equation s = 16t2 . Draw a curve showing
the relation between s and t. Also determine the velocity of the body
at the following times from its being let drop: t = 2 seconds; t = 4.6
seconds; t = 0.01 second.
(3) If x = at − 12 gt2 ; find ẋ and ẍ.
CALCULUS MADE EASY
64
(4) If a body move according to the law
s = 12 − 4.5t + 6.2t2 ,
find its velocity when t = 4 seconds; s being in feet.
(5) Find the acceleration of the body mentioned in the preceding
example. Is the acceleration the same for all values of t?
(6) The angle θ (in radians) turned through by a revolving wheel is
connected with the time t (in seconds) that has elapsed since starting;
by the law
θ = 2.1 − 3.2t + 4.8t2 .
Find the angular velocity (in radians per second) of that wheel when
1 21
seconds have elapsed. Find also its angular acceleration.
(7) A slider moves so that, during the first part of its motion, its
distance s in inches from its starting point is given by the expression
s = 6.8t3 − 10.8t;
t being in seconds.
Find the expression for the velocity and the acceleration at any
time; and hence find the velocity and the acceleration after 3 seconds.
(8) The motion of a rising balloon is such that its height h, in miles,
√
1 3
is given at any instant by the expression h = 0.5 + 10
t − 125; t being
in seconds.
Find an expression for the velocity and the acceleration at any time.
Draw curves to show the variation of height, velocity and acceleration
during the first ten minutes of the ascent.
WHEN TIME VARIES
65
(9) A stone is thrown downwards into water and its depth p in metres at any instant t seconds after reaching the surface of the water is
given by the expression
p=
4
+ 0.8t − 1.
4 + t2
Find an expression for the velocity and the acceleration at any time.
Find the velocity and acceleration after 10 seconds.
(10) A body moves in such a way that the spaces described in the
time t from starting is given by s = tn , where n is a constant. Find the
value of n when the velocity is doubled from the 5th to the 10th second;
find it also when the velocity is numerically equal to the acceleration
at the end of the 10th second.
CHAPTER IX.
INTRODUCING A USEFUL DODGE.
Sometimes one is stumped by finding that the expression to be differentiated is too complicated to tackle directly.
Thus, the equation
3
y = (x2 + a2 ) 2
is awkward to a beginner.
Now the dodge to turn the difficulty is this: Write some symbol,
such as u, for the expression x2 + a2 ; then the equation becomes
3
y = u2 ,
which you can easily manage; for
3 1
dy
= u2 .
du
2
Then tackle the expression
u = x 2 + a2 ,
and differentiate it with respect to x,
du
= 2x.
dx
INTRODUCING A USEFUL DODGE
67
Then all that remains is plain sailing;
dy du
dy
=
× ;
dx
du dx
dy
3 1
= u 2 × 2x
dx
2
1
= 32 (x2 + a2 ) 2 × 2x
for
that is,
1
= 3x(x2 + a2 ) 2 ;
and so the trick is done.
By and bye, when you have learned how to deal with sines, and
cosines, and exponentials, you will find this dodge of increasing usefulness.
Examples.
Let us practise this dodge on a few examples.
√
(1) Differentiate y = a + x.
Let a + x = u.
1
1
1
du
dy
= 1; y = u 2 ;
= 12 u− 2 = 12 (a + x)− 2 .
dx
du
dy
dy du
1
=
×
= √
.
dx
du dx
2 a+x
(2) Differentiate y = √
Let a + x2 = u.
1
.
a + x2
1
3
du
dy
= 2x; y = u− 2 ;
= − 21 u− 2 .
dx
du
dy
dy du
x
=
×
= −p
.
dx
du dx
(a + x2 )3
68
CALCULUS MADE EASY
m − nx +
(3) Differentiate y =
2
2
3
4
p
4
x3
a
.
Let m − nx 3 + px− 3 = u.
1
7
du
= − 23 nx− 3 − 43 px− 3 ;
dx
dy
y = ua ;
= aua−1 .
du
a−1
1
7
2
p
dy
dy du
( 32 nx− 3 + 43 px− 3 ).
=
×
= −a m − nx 3 + 4
dx
du dx
x3
(4) Differentiate y = √
Let u = x3 − a2 .
1
.
− a2
x3
1
3
1
dy
du
= 3x2 ; y = u− 2 ;
= − (x3 − a2 )− 2 .
dx
du
2
2
dy
dy du
3x
=
×
=− p
.
dx
du dx
2 (x3 − a2 )3
r
1−x
(5) Differentiate y =
.
1+x
1
(1 − x) 2
Write this as y =
1 .
(1 + x) 2
1
1
d(1 − x) 2
1 d(1 + x) 2
2
(1
+
x)
−
(1
−
x)
dy
dx
dx
=
.
dx
1+x
1
2
1
1
(We may also write y = (1 − x) 2 (1 + x)− 2 and differentiate as a
product.)
Proceeding as in example (1) above, we get
1
d(1 − x) 2
1
=− √
;
dx
2 1−x
1
and
d(1 + x) 2
1
= √
.
dx
2 1+x
INTRODUCING A USEFUL DODGE
69
Hence
1
1
(1 + x) 2
(1 − x) 2
dy
√
√
=−
−
dx
2(1 + x) 1 − x 2(1 + x) 1 + x
√
1
1−x
√
;
=− √
− p
2 1 + x 1 − x 2 (1 + x)3
dy
1
√
or
=−
.
dx
(1 + x) 1 − x2
r
x3
.
(6) Differentiate y =
1 + x2
We may write this
3
1
y = x 2 (1 + x2 )− 2 ;
1
d (1 + x2 )− 2
3
dy
2 − 12
3 12
= 2 x (1 + x ) + x 2 ×
.
dx
dx
1
Differentiating (1 + x2 )− 2 , as shown in example (2) above, we get
1
d (1 + x2 )− 2
x
= −p
;
dx
(1 + x2 )3
so that
√
√
√
dy
3 x
x5
x(3 + x2 )
= √
−p
= p
.
dx
2 1 + x2
(1 + x2 )3
2 (1 + x2 )3
(7) Differentiate y = (x +
√
Let x + x2 + x + a = u.
y = u3 ;
√
x2 + x + a)3 .
1
d (x2 + x + a) 2
du
=1+
.
dx
dx
2
√
dy
2
2
and
= 3u = 3 x + x + x + a .
du
70
CALCULUS MADE EASY
1
Now let (x2 + x + a) 2 = v and (x2 + x + a) = w.
dw
dx
dv
dx
du
Hence
dx
dy
dx
= 2x + 1;
1
v = w2;
1
dv
= 12 w− 2 .
dw
1
dv
dw
×
= 21 (x2 + x + a)− 2 (2x + 1).
dw
dx
2x + 1
=1+ √
,
2 x2 + x + a
dy du
=
×
du dx
2 √
2x
+
1
= 3 x + x2 + x + a
.
1+ √
2 x2 + x + a
r
r
a2 + x 2 3 a2 − x 2
(8) Differentiate y =
.
a2 − x 2 a2 + x 2
We get
y=
=
1
1
1
2 2
1
2 3
(a2 + x2 ) 2 (a2 − x2 ) 3
1
(a2 − x ) (a2 + x )
2
1
2 − 61
d (a2 + x2 ) 6
dy
2
2 16 d (a − x )
= (a + x )
+
.
1
dx
dx
(a2 − x2 ) 6 dx
1
Let u = (a2 − x2 )− 6 and v = (a2 − x2 ).
du
1 7
dv
= − v− 6 ;
= −2x.
dv
6
dx
7
du
du dv
1
=
×
= x(a2 − x2 )− 6 .
dx
dv dx
3
1
u = v− 6 ;
1
Let w = (a2 + x2 ) 6 and z = (a2 + x2 ).
dw
1 5
dz
= z− 6 ;
= 2x.
dz
6
dx
5
dw
dw dz
1
=
×
= x(a2 + x2 )− 6 .
dx
dz
dx
3
1
w = z6;
1
= (a2 + x2 ) 6 (a2 − x2 )− 6 .
INTRODUCING A USEFUL DODGE
Hence
or
1
x
x
dy
= (a2 + x2 ) 6
7 +
1
5 ;
dx
3(a2 − x2 ) 6
3(a2 − x2 ) 6 (a2 + x2 ) 6
"s
#
x 6 a2 + x 2
1
dy
=
+p
.
6
dx
3
(a2 − x2 )7
(a2 − x2 )(a2 + x2 )5 ]
(9) Differentiate y n with respect to y 5 .
d(y n )
ny n−1
n
=
= y n−5 .
5
5−1
d(y )
5y
5
(10) Find the first and second differential coefficients
xp
of y =
(a − x)x.
b
1
dy
x d (a − x)x 2
=
dx
b
dx
p
(a − x)x
+
.
b
1
1
Let (a − x)x 2 = u and let (a − x)x = w; then u = w 2 .
du
1 1
1
1
p
= w− 2 =
.
1 =
dw
2
2 (a − x)x
2w 2
dw
= a − 2x.
dx
du
dw
du
a − 2x
×
=
= p
.
dw
dx
dx
2 (a − x)x
Hence
x(a − 2x)
dy
= p
+
dx
2b (a − x)x
p
(a − x)x
x(3a − 4x)
= p
.
b
2b (a − x)x
71
72
CALCULUS MADE EASY
Now
p
(3ax − 4x2 )b(a − 2x)
p
(a
−
x)x
(3a
−
8x)
−
2b
(a − x)x
d2 y
=
dx2
4b2 (a − x)x
2
2
3a − 12ax + 8x
p
=
.
4b(a − x) (a − x)x
(We shall need these two last differential coefficients later on. See
Ex. X. No. 11.)
Exercises VI.
(See page 255 for Answers.)
Differentiate the following:
(1) y =
√
(2) y =
1
.
a+x
a
.
(4) y = √
a − x2
(3) y = √
(5) y =
(7) y =
√
x2 + 1.
√
x 2 − a2
.
x2
x 2 + a2 .
√
3 4
x +a
(6) y = √
.
2 3
x +a
a2 + x 2
.
(a + x)2
(8) Differentiate y 5 with respect to y 2 .
√
1 − θ2
(9) Differentiate y =
.
1−θ
The process can be extended to three or more differential coeffidy
dy dz dv
=
×
× .
cients, so that
dx
dz dv dx
INTRODUCING A USEFUL DODGE
Examples.
(1) If z = 3x4 ; v =
We have
73
√
dv
7
; y = 1 + v, find
.
2
z
dx
dv
1
14
dz
dy
;
= √
= − 3;
= 12x3 .
dv
dz
z
dx
2 1+v
dy
168x3
28
√
√
=−
=
−
.
dx
(2 1 + v)z 3
3x5 9x8 + 7
t
1
7x2
dv
(2) If t = √ ; x = t3 + ; v = √
, find
.
3
2
dθ
x−1
5 θ
7x(5x − 6)
dv
= p
;
dx
3 3 (x − 1)4
Hence
dx
= 3t2 + 21 ;
dt
dt
1
=− √ .
dθ
10 θ3
7x(5x − 6)(3t2 + 21 )
dv
√
p
=−
,
dθ
30 3 (x − 1)4 θ3
an expression in which x must be replaced by its value, and t by its
value in terms of θ.
√
√
3a2 x
1 − θ2
1
dφ
(3) If θ = √ ; ω =
; and φ = 3 − √ , find
.
1+θ
dx
ω 2
x3
We get
r
√
1−θ
1
2 − 21
θ = 3a x ; ω =
; and φ = 3 − √ ω −1 .
1+θ
2
2
dθ
3a
dω
1
√
=− √ ;
=−
3
dx
dθ
(1 + θ) 1 − θ2
2 x
(see example 5, p. 68); and
dφ
1
=√
.
dω
2ω 2
1
3a2
dθ
1
√
=√
×
× √ .
dx
2 × ω 2 (1 + θ) 1 − θ2 2 x3
Replace now first ω, then θ by its value.
So that
CALCULUS MADE EASY
Exercises VII.
74
You can now successfully try the following. (See
page 256 for Answers.)
dw
1
, find
.
2
v
dx
√
√
1
dv
(2) If y = 3x2 + 2; z = 1 + y; and v = √
, find
.
dx
3 + 4z
x3
1
du
(3) If y = √ ; z = (1 + y)2 ; and u = √
, find
.
dx
1+z
3
(1) If u = 21 x3 ; v = 3(u + u2 ); and w =
CHAPTER X.
GEOMETRICAL MEANING OF DIFFERENTIATION.
It is useful to consider what geometrical meaning can be given to the
differential coefficient.
In the first place, any function of x, such, for example, as x2 , or
√
x,
or ax + b, can be plotted as a curve; and nowadays every schoolboy is
familiar with the process of curve-plotting.
R
Y
Q
dy
dx
P
y
O
x
dx
X
Fig. 7.
Let P QR, in Fig. 7, be a portion of a curve plotted with respect
to the axes of coordinates OX and OY . Consider any point Q on this
curve, where the abscissa of the point is x and its ordinate is y. Now
observe how y changes when x is varied. If x is made to increase by
CALCULUS MADE EASY
76
a small increment dx, to the right, it will be observed that y also (in
this particular curve) increases by a small increment dy (because this
particular curve happens to be an ascending curve). Then the ratio of
dy to dx is a measure of the degree to which the curve is sloping up
between the two points Q and T . As a matter of fact, it can be seen on
the figure that the curve between Q and T has many different slopes,
so that we cannot very well speak of the slope of the curve between
Q and T . If, however, Q and T are so near each other that the small
portion QT of the curve is practically straight, then it is true to say that
dy
is the slope of the curve along QT . The straight line QT
the ratio
dx
produced on either side touches the curve along the portion QT only,
and if this portion is indefinitely small, the straight line will touch the
curve at practically one point only, and be therefore a tangent to the
curve.
This tangent to the curve has evidently the same slope as QT , so
dy
that
is the slope of the tangent to the curve at the point Q for which
dx
dy
the value of
is found.
dx
We have seen that the short expression “the slope of a curve” has
no precise meaning, because a curve has so many slopes—in fact, every
small portion of a curve has a different slope. “The slope of a curve at
a point” is, however, a perfectly defined thing; it is the slope of a very
small portion of the curve situated just at that point; and we have seen
that this is the same as “the slope of the tangent to the curve at that
point.”
Observe that dx is a short step to the right, and dy the corresponding short step upwards. These steps must be considered as short as
MEANING OF DIFFERENTIATION
77
possible—in fact indefinitely short,—though in diagrams we have to
represent them by bits that are not infinitesimally small, otherwise
they could not be seen.
We shall hereafter make considerable use of this circumstance that
dy
represents the slope of the curve at any point.
dx
Y
dy
dx
O
X
Fig. 8.
If a curve is sloping up at 45◦ at a particular point, as in Fig. 8, dy
dy
and dx will be equal, and the value of
= 1.
dx
dy
If the curve slopes up steeper than 45◦ (Fig. 9),
will be greater
dx
than 1.
dy
If the curve slopes up very gently, as in Fig. 10,
will be a fraction
dx
smaller than 1.
For a horizontal line, or a horizontal place in a curve, dy = 0, and
dy
therefore
= 0.
dx
If a curve slopes downward, as in Fig. 11, dy will be a step down,
dy
will have
and must therefore be reckoned of negative value; hence
dx
negative sign also.
78
CALCULUS MADE EASY
Y
Y
dy
dy
dx
dx
O
X
O
X
Fig. 9.
Fig. 10.
If the “curve” happens to be a straight line, like that in Fig. 12, the
dy
will be the same at all points along it. In other words its
value of
dx
slope is constant.
If a curve is one that turns more upwards as it goes along to the
dy
right, the values of
will become greater and greater with the indx
creasing steepness, as in Fig. 13.
If a curve is one that gets flatter and flatter as it goes along, the
dy
values of
will become smaller and smaller as the flatter part is
dx
Y
Q dx
dy
y
O
x
dx
Fig. 11.
X
MEANING OF DIFFERENTIATION
79
Y
dy
dx
dy
dx
dy
dx
O
X
Fig. 12.
Y
dy
dx
dy
dy
dx
dx
O
X
Fig. 13.
reached, as in Fig. 14.
If a curve first descends, and then goes up again, as in Fig. 15,
dy
presenting a concavity upwards, then clearly
will first be negative,
dx
with diminishing values as the curve flattens, then will be zero at the
point where the bottom of the trough of the curve is reached; and from
dy
this point onward
will have positive values that go on increasing. In
dx
such a case y is said to pass by a minimum. The minimum value of y is
not necessarily the smallest value of y, it is that value of y corresponding
to the bottom of the trough; for instance, in Fig. 28 (p. 99), the value
of y corresponding to the bottom of the trough is 1, while y takes
80
CALCULUS MADE EASY
Y
Y
y min.
O
X
Fig. 14.
O
X
Fig. 15.
elsewhere values which are smaller than this. The characteristic of a
minimum is that y must increase on either side of it.
N.B.—For the particular value of x that makes y a minimum, the
dy
= 0.
value of
dx
dy
If a curve first ascends and then descends, the values of
will be
dx
positive at first; then zero, as the summit is reached; then negative,
as the curve slopes downwards, as in Fig. 16. In this case y is said to
pass by a maximum, but the maximum value of y is not necessarily the
greatest value of y. In Fig. 28, the maximum of y is 2 13 , but this is
by no means the greatest value y can have at some other point of the
curve.
N.B.—For the particular value of x that makes y a maximum, the
dy
value of
= 0.
dx
dy
will
If a curve has the peculiar form of Fig. 17, the values of
dx
always be positive; but there will be one particular place where the
dy
slope is least steep, where the value of
will be a minimum; that is,
dx
less than it is at any other part of the curve.
MEANING OF DIFFERENTIATION
Y
81
Y
y max.
O
X
O
X
Fig. 16.
Fig. 17.
dy
will be negative
dx
in the upper part, and positive in the lower part; while at the nose of
dy
the curve where it becomes actually perpendicular, the value of
will
dx
be infinitely great.
If a curve has the form of Fig. 18, the value of
Y
dx
dy
Q
dy
dx
O
X
Fig. 18.
dy
measures the steepness of a curve
dx
at any point, let us turn to some of the equations which we have already
Now that we understand that
learned how to differentiate.
82
CALCULUS MADE EASY
(1) As the simplest case take this:
y = x + b.
It is plotted out in Fig. 19, using equal scales for x and y. If we put
x = 0, then the corresponding ordinate will be y = b; that is to say, the
“curve” crosses the y-axis at the height b. From here it ascends at 45◦ ;
Y
Y
dy
dx
b
b
O
X
O
Fig. 19.
X
Fig. 20.
for whatever values we give to x to the right, we have an equal y to
ascend. The line has a gradient of 1 in 1.
Now differentiate y = x + b, by the rules we have already learned
dy
(pp. 21 and 25 ante), and we get
= 1.
dx
The slope of the line is such that for every little step dx to the right,
we go an equal little step dy upward. And this slope is constant—always
the same slope.
(2) Take another case:
y = ax + b.
MEANING OF DIFFERENTIATION
83
We know that this curve, like the preceding one, will start from a
height b on the y-axis. But before we draw the curve, let us find its
dy
= a. The slope will be constant,
slope by differentiating; which gives
dx
at an angle, the tangent of which is here called a. Let us assign to a
some numerical value—say 13 . Then we must give it such a slope that
it ascends 1 in 3; or dx will be 3 times as great as dy; as magnified in
Fig. 21. So, draw the line in Fig. 20 at this slope.
Fig. 21.
(3) Now for a slightly harder case.
Let
y = ax2 + b.
Again the curve will start on the y-axis at a height b above the
origin.
Now differentiate. [If you have forgotten, turn back to p. 25; or,
rather, don’t turn back, but think out the differentiation.]
dy
= 2ax.
dx
This shows that the steepness will not be constant: it increases as
x increases. At the starting point P , where x = 0, the curve (Fig. 22)
has no steepness—that is, it is level. On the left of the origin, where x
dy
has negative values,
will also have negative values, or will descend
dx
from left to right, as in the Figure.
84
CALCULUS MADE EASY
Y
R
Q
P
b
O
X
Fig. 22.
Let us illustrate this by working out a particular instance. Taking
the equation
y = 14 x2 + 3,
and differentiating it, we get
dy
= 12 x.
dx
Now assign a few successive values, say from 0 to 5, to x; and calculate
dy
from the
the corresponding values of y by the first equation; and of
dx
second equation. Tabulating results, we have:
x
0
1
2
3
4
5
y
3
3 14
4
5 14
7
9 14
dy
dx
0
1
2
1
1 12
2
2 12
Then plot them out in two curves, Figs. 23 and 24, in Fig. 23 plotting
dy
the values of y against those of x and in Fig. 24 those of
against
dx
those of x. For any assigned value of x, the height of the ordinate in
the second curve is proportional to the slope of the first curve.
85
MEANING OF DIFFERENTIATION
y
dy
dx
x2
+
3
9
8
7
6
5
y
=
4
1
1 x2
4
b
−3 −2 −1 0
1
2
3
4
5
x
5
4
3
2
−3 −2 −1 1
dy = 12x
dx
0
Fig. 23.
1
2
3
4
5
Fig. 24.
If a curve comes to a sudden cusp, as in Fig. 25, the slope at that
point suddenly changes from a slope upward to a slope downward. In
Y
O
X
Fig. 25.
dy
will clearly undergo an abrupt change from a positive to
dx
a negative value.
that case
The following examples show further applications of the principles
just explained.
x
CALCULUS MADE EASY
86
(4) Find the slope of the tangent to the curve
y=
1
+ 3,
2x
at the point where x = −1. Find the angle which this tangent makes
with the curve y = 2x2 + 2.
The slope of the tangent is the slope of the curve at the point where
dy
of the curve for
they touch one another (see p. 76); that is, it is the
dx
dy
1
dy
1
that point. Here
= − 2 and for x = −1,
= − , which is the
dx
2x
dx
2
slope of the tangent and of the curve at that point. The tangent, being
dy
= a,
a straight line, has for equation y = ax + b, and its slope is
dx
1
1
hence a = − . Also if x = −1, y =
+ 3 = 2 12 ; and as the
2
2(−1)
tangent passes by this point, the coordinates of the point must satisfy
the equation of the tangent, namely
1
y = − x + b,
2
1
so that 2 21 = − × (−1) + b and b = 2; the equation of the tangent is
2
1
therefore y = − x + 2.
2
Now, when two curves meet, the intersection being a point common to both curves, its coordinates must satisfy the equation of each
one of the two curves; that is, it must be a solution of the system of
simultaneous equations formed by coupling together the equations of
the curves. Here the curves meet one another at points given by the
solution of

 y = 2x2 + 2,
 y = − 1 x + 2 or 2x2 + 2 = − 1 x + 2;
2
2
87
MEANING OF DIFFERENTIATION
that is,
x(2x + 21 ) = 0.
This equation has for its solutions x = 0 and x = − 41 . The slope of
the curve y = 2x2 + 2 at any point is
dy
= 4x.
dx
For the point where x = 0, this slope is zero; the curve is horizontal.
For the point where
dy
1
= −1;
x=− ,
4
dx
hence the curve at that point slopes downwards to the right at such
an angle θ with the horizontal that tan θ = 1; that is, at 45◦ to the
horizontal.
The slope of the straight line is − 21 ; that is, it slopes downwards to
the right and makes with the horizontal an angle φ such that tan φ = 12 ;
that is, an angle of 26◦ 340 . It follows that at the first point the curve
cuts the straight line at an angle of 26◦ 340 , while at the second it cuts
it at an angle of 45◦ − 26◦ 340 = 18◦ 260 .
(5) A straight line is to be drawn, through a point whose coordinates
are x = 2, y = −1, as tangent to the curve y = x2 − 5x + 6. Find the
coordinates of the point of contact.
The slope of the tangent must be the same as the
that is, 2x − 5.
dy
of the curve;
dx
The equation of the straight line is y = ax+b, and as it is satisfied for
dy
the values x = 2, y = −1, then −1 = a×2+b; also, its
= a = 2x−5.
dx
The x and the y of the point of contact must also satisfy both the
equation of the tangent and the equation of the curve.
CALCULUS MADE EASY
88
We have then








y = x2 − 5x + 6,
(i)
y = ax + b,
(ii)


−1 = 2a + b,





a = 2x − 5,
(iii)
(iv)
four equations in a, b, x, y.
Equations (i) and (ii) give x2 − 5x + 6 = ax + b.
Replacing a and b by their value in this, we get
x2 − 5x + 6 = (2x − 5)x − 1 − 2(2x − 5),
which simplifies to x2 − 4x + 3 = 0, the solutions of which are: x = 3
and x = 1. Replacing in (i), we get y = 0 and y = 2 respectively; the
two points of contact are then x = 1, y = 2, and x = 3, y = 0.
Note.—In all exercises dealing with curves, students will find it extremely instructive to verify the deductions obtained by actually plotting the curves.
Exercises VIII.
(See page 256 for Answers.)
(1) Plot the curve y = 43 x2 −5, using a scale of millimetres. Measure
at points corresponding to different values of x, the angle of its slope.
Find, by differentiating the equation, the expression for slope; and
see, from a Table of Natural Tangents, whether this agrees with the
measured angle.
MEANING OF DIFFERENTIATION
89
(2) Find what will be the slope of the curve
y = 0.12x3 − 2,
at the particular point that has as abscissa x = 2.
(3) If y = (x − a)(x − b), show that at the particular point of the
dy
= 0, x will have the value 12 (a + b).
curve where
dx
dy
of the equation y = x3 + 3x; and calculate the
(4) Find the
dx
dy
numerical values of
for the points corresponding to x = 0, x = 12 ,
dx
x = 1, x = 2.
(5) In the curve to which the equation is x2 + y 2 = 4, find the values
of x at those points where the slope = 1.
(6) Find the slope, at any point, of the curve whose equation is
y2
x2
+
= 1; and give the numerical value of the slope at the place
32
22
where x = 0, and at that where x = 1.
(7) The equation of a tangent to the curve y = 5 − 2x + 0.5x3 , being
of the form y = mx + n, where m and n are constants, find the value
of m and n if the point where the tangent touches the curve has x = 2
for abscissa.
(8) At what angle do the two curves
y = 3.5x2 + 2 and y = x2 − 5x + 9.5
cut one another?
√
(9) Tangents to the curve y = ± 25 − x2 are drawn at points for
which x = 3 and x = 4. Find the coordinates of the point of intersection
of the tangents and their mutual inclination.
CALCULUS MADE EASY
90
(10) A straight line y = 2x − b touches a curve y = 3x2 + 2 at one
point. What are the coordinates of the point of contact, and what is
the value of b?
CHAPTER XI.
MAXIMA AND MINIMA.
One of the principal uses of the process of differentiating is to find out
under what conditions the value of the thing differentiated becomes a
maximum, or a minimum. This is often exceedingly important in engineering questions, where it is most desirable to know what conditions
will make the cost of working a minimum, or will make the efficiency a
maximum.
Now, to begin with a concrete case, let us take the equation
y = x2 − 4x + 7.
By assigning a number of successive values to x, and finding the corresponding values of y, we can readily see that the equation represents
a curve with a minimum.
x
0
1
2
3
4
5
y
7
4
3
4
7
12
These values are plotted in Fig. 26, which shows that y has apparently a minimum value of 3, when x is made equal to 2. But are you
sure that the minimum occurs at 2, and not at 2 14 or at 1 34 ?
92
CALCULUS MADE EASY
Y
7
6
5
4
3
2
1
O
1
2
3
X
4
Fig. 26.
Of course it would be possible with any algebraic expression to work
out a lot of values, and in this way arrive gradually at the particular
value that may be a maximum or a minimum.
Y
4
3
2
1
−1
O
−1
−2
−3
−4
1
2
3
Fig. 27.
Here is another example:
Let
y = 3x − x2 .
4
X
93
MAXIMA AND MINIMA
Calculate a few values thus:
x
−1
0
1
2
3
4
5
y
−4
0
2
2
0
−4
−10
Plot these values as in Fig. 27.
It will be evident that there will be a maximum somewhere between
x = 1 and x = 2; and the thing looks as if the maximum value of y ought
to be about 2 14 . Try some intermediate values. If x = 1 41 , y = 2.187; if
x = 1 12 , y = 2.25; if x = 1.6, y = 2.24. How can we be sure that 2.25 is
the real maximum, or that it occurs exactly when x = 1 21 ?
Now it may sound like juggling to be assured that there is a way
by which one can arrive straight at a maximum (or minimum) value
without making a lot of preliminary trials or guesses. And that way
depends on differentiating. Look back to an earlier page (78) for the
remarks about Figs. 14 and 15, and you will see that whenever a curve
gets either to its maximum or to its minimum height, at that point its
dy
= 0. Now this gives us the clue to the dodge that is wanted. When
dx
there is put before you an equation, and you want to find that value
of x that will make its y a minimum (or a maximum), first differentiate
dy
it, and having done so, write its
as equal to zero, and then solve
dx
for x. Put this particular value of x into the original equation, and you
will then get the required value of y. This process is commonly called
“equating to zero.”
To see how simply it works, take the example with which this chapter opens, namely
y = x2 − 4x + 7.
CALCULUS MADE EASY
94
Differentiating, we get:
dy
= 2x − 4.
dx
Now equate this to zero, thus:
2x − 4 = 0.
Solving this equation for x, we get:
2x = 4,
x = 2.
Now, we know that the maximum (or minimum) will occur exactly
when x = 2.
Putting the value x = 2 into the original equation, we get
y = 22 − (4 × 2) + 7
=4−8+7
= 3.
Now look back at Fig. 26, and you will see that the minimum occurs
when x = 2, and that this minimum of y = 3.
Try the second example (Fig. 24), which is
y = 3x − x2 .
Differentiating,
dy
= 3 − 2x.
dx
Equating to zero,
3 − 2x = 0,
whence
x = 1 21 ;
MAXIMA AND MINIMA
95
and putting this value of x into the original equation, we find:
y = 4 21 − (1 12 × 1 12 ),
y = 2 14 .
This gives us exactly the information as to which the method of trying
a lot of values left us uncertain.
Now, before we go on to any further cases, we have two remarks to
dy
to zero, you feel at first (that
make. When you are told to equate
dx
is if you have any wits of your own) a kind of resentment, because you
dy
has all sorts of different values at different parts of the
know that
dx
curve, according to whether it is sloping up or down. So, when you are
suddenly told to write
dy
= 0,
dx
you resent it, and feel inclined to say that it can’t be true. Now you will
have to understand the essential difference between “an equation,” and
“an equation of condition.” Ordinarily you are dealing with equations
that are true in themselves, but, on occasions, of which the present are
examples, you have to write down equations that are not necessarily
true, but are only true if certain conditions are to be fulfilled; and you
write them down in order, by solving them, to find the conditions which
make them true. Now we want to find the particular value that x has
when the curve is neither sloping up nor sloping down, that is, at the
dy
dy
particular place where
= 0. So, writing
= 0 does not mean that
dx
dx
it always is = 0; but you write it down as a condition in order to see
dy
how much x will come out if
is to be zero.
dx
CALCULUS MADE EASY
96
The second remark is one which (if you have any wits of your own)
you will probably have already made: namely, that this much-belauded
process of equating to zero entirely fails to tell you whether the x that
you thereby find is going to give you a maximum value of y or a minimum value of y. Quite so. It does not of itself discriminate; it finds
for you the right value of x but leaves you to find out for yourselves
whether the corresponding y is a maximum or a minimum. Of course,
if you have plotted the curve, you know already which it will be.
For instance, take the equation:
1
y = 4x + .
x
Without stopping to think what curve it corresponds to, differentiate it, and equate to zero:
whence
1
dy
= 4 − x−2 = 4 − 2 = 0;
dx
x
1
x = 2;
and, inserting this value,
y=4
will be either a maximum or else a minimum. But which? You will
hereafter be told a way, depending upon a second differentiation, (see
Chap. XII., p. 109). But at present it is enough if you will simply
try any other value of x differing a little from the one found, and see
whether with this altered value the corresponding value of y is less or
greater than that already found.
MAXIMA AND MINIMA
97
Try another simple problem in maxima and minima. Suppose you
were asked to divide any number into two parts, such that the product
was a maximum? How would you set about it if you did not know the
trick of equating to zero? I suppose you could worry it out by the rule
of try, try, try again. Let 60 be the number. You can try cutting it into
two parts, and multiplying them together. Thus, 50 times 10 is 500; 52
times 8 is 416; 40 times 20 is 800; 45 times 15 is 675; 30 times 30 is 900.
This looks like a maximum: try varying it. 31 times 29 is 899, which is
not so good; and 32 times 28 is 896, which is worse. So it seems that
the biggest product will be got by dividing into two equal halves.
Now see what the calculus tells you. Let the number to be cut into
two parts be called n. Then if x is one part, the other will be n − x,
and the product will be x(n − x) or nx − x2 . So we write y = nx − x2 .
Now differentiate and equate to zero;
Solving for x, we get
dy
= n − 2x = 0
dx
n
= x.
2
So now we know that whatever number n may be, we must divide it
into two equal parts if the product of the parts is to be a maximum;
and the value of that maximum product will always be = 41 n2 .
This is a very useful rule, and applies to any number of factors, so
that if m + n + p = a constant number, m × n × p is a maximum when
m = n = p.
CALCULUS MADE EASY
98
Test Case.
Let us at once apply our knowledge to a case that we can test.
y = x2 − x;
Let
and let us find whether this function has a maximum or minimum; and
if so, test whether it is a maximum or a minimum.
Differentiating, we get
dy
= 2x − 1.
dx
Equating to zero, we get
2x − 1 = 0,
whence
2x = 1,
x = 21 .
or
That is to say, when x is made = 21 , the corresponding value of y
will be either a maximum or a minimum. Accordingly, putting x =
1
2
in the original equation, we get
y = ( 21 )2 − 12 ,
or
y = − 14 .
Is this a maximum or a minimum? To test it, try putting x a little
bigger than 21 ,—say make x = 0.6. Then
y = (0.6)2 − 0.6 = 0.36 − 0.6 = −0.24,
which is higher up than −0.25; showing that y = −0.25 is a minimum.
Plot the curve for yourself, and verify the calculation.
99
MAXIMA AND MINIMA
Further Examples.
A most interesting example is afforded by a curve that has both a
maximum and a minimum. Its equation is:
y = 31 x3 − 2x2 + 3x + 1.
dy
= x2 − 4x + 3.
dx
Now
Y
6
5
4
3
2
−1
1
1
O
−1
2
3
4
5
−2
−3
−4
Fig. 28.
Equating to zero, we get the quadratic,
x2 − 4x + 3 = 0;
and solving the quadratic gives us two roots, viz.

x = 3
 x = 1.
X
100
CALCULUS MADE EASY
Now, when x = 3, y = 1; and when x = 1, y = 2 13 . The first of
these is a minimum, the second a maximum.
The curve itself may be plotted (as in Fig. 28) from the values
calculated, as below, from the original equation.
x
−1
0
1
2
3
4
5
6
y
−4 13
1
2 13
1 32
1
2 13
7 23
19
A further exercise in maxima and minima is afforded by the following example:
The equation to a circle of radius r, having its centre C at the point
whose coordinates are x = a, y = b, as depicted in Fig. 29, is:
(y − b)2 + (x − a)2 = r2 .
Y
r
C
y
b
x
O
X
a
Fig. 29.
This may be transformed into
y=
p
r2 − (x − a)2 + b.
MAXIMA AND MINIMA
101
Now we know beforehand, by mere inspection of the figure, that
when x = a, y will be either at its maximum value, b + r, or else at its
minimum value, b − r. But let us not take advantage of this knowledge;
let us set about finding what value of x will make y a maximum or a
minimum, by the process of differentiating and equating to zero.
1
1
dy
= p
× (2a − 2x),
2
dx
2 r − (x − a)2
which reduces to
a−x
dy
=p
.
dx
r2 − (x − a)2
Then the condition for y being maximum or minimum is:
a−x
p
= 0.
r2 − (x − a)2
Since no value whatever of x will make the denominator infinite,
the only condition to give zero is
x = a.
Inserting this value in the original equation for the circle, we find
y=
√
r2 + b;
and as the root of r2 is either +r or −r, we have two resulting values
of y,

y = b + r
 y = b − r.
CALCULUS MADE EASY
102
The first of these is the maximum, at the top; the second the minimum, at the bottom.
If the curve is such that there is no place that is a maximum or
minimum, the process of equating to zero will yield an impossible result.
For instance:
y = ax3 + bx + c.
Let
dy
= 3ax2 + b.
dx
Equating this to zero, we get 3ax2 + b = 0,
r
−b
−b
2
x =
, and x =
, which is impossible.
3a
3a
Therefore y has no maximum nor minimum.
Then
A few more worked examples will enable you to thoroughly master
this most interesting and useful application of the calculus.
(1) What are the sides of the rectangle of maximum area inscribed
in a circle of radius R?
If one side be called x,
the other side =
p
(diagonal)2 − x2 ;
and as the diagonal of the rectangle is necessarily a diameter, the other
√
side = 4R2 − x2 .
√
Then, area of rectangle S = x 4R2 − x2 ,
√
√
d 4R2 − x2
dS
d(x)
=x×
+ 4R2 − x2 ×
.
dx
dx
dx
√
If you have forgotten how to differentiate 4R2 − x2 , here is a hint:
√
dy
dw
and
; fight it out,
write 4R2 − x2 = w and y = w, and seek
dw
dx
and only if you can’t get on refer to page 66.
103
MAXIMA AND MINIMA
You will get
√
x
4R2 − 2x2
dS
= x × −√
+ 4R2 − x2 = √
.
dx
4R2 − x2
4R2 − x2
For maximum or minimum we must have
4R2 − 2x2
√
= 0;
4R2 − x2
√
that is, 4R2 − 2x2 = 0 and x = R 2.
√
√
The other side = 4R2 − 2R2 = R 2; the two sides are equal; the
figure is a square the side of which is equal to the diagonal of the square
constructed on the radius. In this case it is, of course, a maximum with
which we are dealing.
(2) What is the radius of the opening of a conical vessel the sloping
side of which has a length l when the capacity of the vessel is greatest?
√
If R be the radius and H the corresponding height, H = l2 − R2 .
√
H
l 2 − R2
2
2
Volume V = πR ×
= πR ×
.
3
3
Proceeding as in the previous problem, we get
dV
R
2πR √ 2
= πR2 × − √
+
l − R2
dR
3
3 l 2 − R2
2πR(l2 − R2 ) − πR3
√
=
=0
3 l 2 − R2
for maximum or minimum.
2
2
2
Or, 2πR(l − R ) − πR = 0, and R = l
q
2
,
3
obviously.
(3) Find the maxima and minima of the function
y=
4−x
x
+
.
4−x
x
for a maximum,
CALCULUS MADE EASY
104
We get
(4 − x) − (−x) −x − (4 − x)
dy
=
+
=0
dx
(4 − x)2
x2
for maximum or minimum; or
4
4
− 2 = 0 and x = 2.
2
(4 − x)
x
There is only one value, hence only one maximum or minimum.
For x = 2,
y = 2,
for x = 1.5,
y = 2.27,
for x = 2.5,
y = 2.27;
it is therefore a minimum. (It is instructive to plot the graph of the
function.)
√
(4) Find the maxima and minima of the function y =
1 − x. (It will be found instructive to plot the graph.)
Differentiating gives at once (see example No. 1, p. 67)
dy
1
1
= √
− √
=0
dx
2 1+x 2 1−x
for maximum or minimum.
√
√
Hence 1 + x = 1 − x and x = 0, the only solution
For x = 0, y = 2.
For x = ±0.5, y = 1.932, so this is a maximum.
(5) Find the maxima and minima of the function
y=
x2 − 5
.
2x − 4
√
1+x +
105
MAXIMA AND MINIMA
We have
(2x − 4) × 2x − (x2 − 5)2
dy
=
=0
dx
(2x − 4)2
for maximum or minimum; or
2x2 − 8x + 10
= 0;
(2x − 4)2
or x2 − 4x + 5 = 0; which has for solutions
x=
5
2
±
√
−1.
These being imaginary, there is no real value of x for which
hence there is neither maximum nor minimum.
dy
= 0;
dx
(6) Find the maxima and minima of the function
(y − x2 )2 = x5 .
5
This may be written y = x2 ± x 2 .
3
dy
= 2x ± 25 x 2 = 0 for maximum or minimum;
dx
1
1
that is, x(2± 25 x 2 ) = 0, which is satisfied for x = 0, and for 2± 52 x 2 = 0,
that is for x =
16
.
25
So there are two solutions.
p
Taking first x = 0. If x = −0.5, y = 0.25 ± 2 −(.5)5 , and if
p
x = +0.5, y = 0.25 ± 2 (.5)5 . On one side y is imaginary; that is,
there is no value of y that can be represented by a graph; the latter is
therefore entirely on the right side of the axis of y (see Fig. 30).
On plotting the graph it will be found that the curve goes to the
origin, as if there were a minimum there; but instead of continuing
beyond, as it should do for a minimum, it retraces its steps (forming
106
CALCULUS MADE EASY
what is called a “cusp”). There is no minimum, therefore, although the
dy
= 0. It is necessary
condition for a minimum is satisfied, namely
dx
therefore always to check by taking one value on either side.
y
0.3
0.2
0.1
0
0.2
0.4
0.6
0.8 1.0
x
Fig. 30.
Now, if we take x =
16
25
= 0.64. If x = 0.64, y = 0.7373 and
y = 0.0819; if x = 0.6, y becomes 0.6389 and 0.0811; and if x = 0.7,
y becomes 0.8996 and 0.0804.
This shows that there are two branches of the curve; the upper one
does not pass through a maximum, but the lower one does.
(7) A cylinder whose height is twice the radius of the base is increasing in volume, so that all its parts keep always in the same proportion
to each other; that is, at any instant, the cylinder is similar to the
original cylinder. When the radius of the base is r feet, the surface
area is increasing at the rate of 20 square inches per second; at what
MAXIMA AND MINIMA
107
rate is its volume then increasing?
Area = S = 2(πr2 ) + 2πr × 2r = 6πr2 .
Volume = V = πr2 × 2r = 2πr3 .
dS
= 12πr,
dr
dV
= 6πr2 ,
dr
dS = 12πr dr = 20,
dr =
dV = 6πr2 dr = 6πr2 ×
20
,
12πr
20
= 10r.
12πr
The volume changes at the rate of 10r cubic inches.
Make other examples for yourself. There are few subjects which
offer such a wealth for interesting examples.
Exercises IX.
(See page 257 for Answers.)
(1) What values of x will make y a maximum and a minimum, if
x2
?
y=
x+1
(2) What value of x will make y a maximum in the equation y =
x
?
2
a + x2
(3) A line of length p is to be cut up into 4 parts and put together
as a rectangle. Show that the area of the rectangle will be a maximum
if each of its sides is equal to 14 p.
(4) A piece of string 30 inches long has its two ends joined together
and is stretched by 3 pegs so as to form a triangle. What is the largest
triangular area that can be enclosed by the string?
CALCULUS MADE EASY
108
(5) Plot the curve corresponding to the equation
y=
10
10
+
;
x
8−x
dy
, and deduce the value of x that will make y a minimum;
dx
and find that minimum value of y.
also find
(6) If y = x5 − 5x, find what values of x will make y a maximum or
a minimum.
(7) What is the smallest square that can be inscribed in a given
square?
(8) Inscribe in a given cone, the height of which is equal to the radius
of the base, a cylinder (a) whose volume is a maximum; (b) whose
lateral area is a maximum; (c) whose total area is a maximum.
(9) Inscribe in a sphere, a cylinder (a) whose volume is a maximum; (b) whose lateral area is a maximum; (c) whose total area is a
maximum.
(10) A spherical balloon is increasing in volume. If, when its radius
is r feet, its volume is increasing at the rate of 4 cubic feet per second,
at what rate is its surface then increasing?
(11) Inscribe in a given sphere a cone whose volume is a maximum.
(12) The current C given by a battery of N similar voltaic cells is
n×E
C =
, where E, R, r, are constants and n is the number of
rn2
R+
N
cells coupled in series. Find the proportion of n to N for which the
current is greatest.
CHAPTER XII.
CURVATURE OF CURVES.
Returning to the process of successive differentiation, it may be
asked: Why does anybody want to differentiate twice over? We know
that when the variable quantities are space and time, by differentiating
twice over we get the acceleration of a moving body, and that in the
dy
means the slope of
geometrical interpretation, as applied to curves,
dx
2
dy
mean in this case? Clearly it means the
the curve. But what can
dx2
rate (per unit of length x) at which the slope is changing—in brief, it
is a measure of the curvature of the slope.
Y
Y
O
X
O
Fig. 31.
Suppose a slope constant, as in Fig. 31.
X
Fig. 32.
110
CALCULUS MADE EASY
dy
is of constant value.
dx
Suppose, however, a case in which,
like Fig. 32, the slope itself is
dy
d
d2 y
dx
getting greater upwards, then
, that is, 2 , will be positive.
dx
dx
If the slope is becoming less as you go to the right (as in Fig. 14,
Here,
p. 80), or as in Fig. 33, then, even though the curve may be going
d2 y
will
upward, since the change is such as to diminish its slope, its
dx2
be negative.
Y
O
X
Fig. 33.
It is now time to initiate you into another secret—how to tell
whether the result that you get by “equating to zero” is a maximum or
a minimum. The trick is this: After you have differentiated (so as to
get the expression which you equate to zero), you then differentiate a
second time, and look whether the result of the second differentiation
d2 y
is positive or negative. If
comes out positive, then you know that
dx2
d2 y
the value of y which you got was a minimum; but if
comes out
dx2
negative, then the value of y which you got must be a maximum. That’s
111
CURVATURE OF CURVES
the rule.
The reason of it ought to be quite evident. Think of any curve that
has a minimum point in it (like Fig. 15, p. 80), or like Fig. 34, where the
point of minimum y is marked M , and the curve is concave upwards.
To the left of M the slope is downward, that is, negative, and is getting
less negative. To the right of M the slope has become upward, and
is getting more and more upward. Clearly the change of slope as the
Y
Y
M
M
y max.
y min.
O
X
x
Fig. 34.
O
X
x
Fig. 35.
d2 y
curve passes through M is such that
is positive, for its operation,
dx2
as x increases toward the right, is to convert a downward slope into an
upward one.
Similarly, consider any curve that has a maximum point in it (like
Fig. 16, p. 81), or like Fig. 35, where the curve is convex, and the maximum point is marked M . In this case, as the curve passes through M
from left to right, its upward slope is converted into a downward or
d2 y
negative slope, so that in this case the “slope of the slope”
is
dx2
negative.
Go back now to the examples of the last chapter and verify in this
CALCULUS MADE EASY
112
way the conclusions arrived at as to whether in any particular case
there is a maximum or a minimum. You will find below a few worked
out examples.
(1) Find the maximum or minimum of
(a) y = 4x2 − 9x − 6;
(b) y = 6 + 9x − 4x2 ;
and ascertain if it be a maximum or a minimum in each case.
dy
dx
d2 y
dx2
dy
(b)
dx
d2 y
dx2
(a)
= 8x − 9 = 0;
= 8;
and y = −11.065.
it is +; hence it is a minimum.
= 9 − 8x = 0;
= −8;
x = 1 81 ,
x = 1 81 ;
and y = +11.065.
it is −; hence it is a maximum.
(2) Find the maxima and minima of the function y = x3 − 3x + 16.
dy
= 3x2 − 3 = 0; x2 = 1; and x = ±1.
dx
d2 y
= 6x; for x = 1; it is +;
dx2
hence x = 1 corresponds to a minimum y = 14. For x = −1 it is −;
hence x = −1 corresponds to a maximum y = +18.
x−1
(3) Find the maxima and minima of y = 2
.
x +2
dy
(x2 + 2) × 1 − (x − 1) × 2x
2x − x2 + 2
=
=
= 0;
dx
(x2 + 2)2
(x2 + 2)2
113
CURVATURE OF CURVES
or x2 − 2x − 2 = 0, whose solutions are x = +2.73 and x = −0.73.
(x2 + 2)2 × (2x − 2) − (x2 − 2x − 2)(4x3 + 8x)
d2 y
=
−
dx2
(x2 + 2)4
5
4
3
2x − 6x − 8x − 8x2 − 24x + 8
=−
.
(x2 + 2)4
The denominator is always positive, so it is sufficient to ascertain
the sign of the numerator.
If we put x = 2.73, the numerator is negative; the maximum, y =
0.183.
If we put x = −0.73, the numerator is positive; the minimum,
y = −0.683.
(4) The expense C of handling the products of a certain factory
varies with the weekly output P according to the relation C = aP +
b
+ d, where a, b, c, d are positive constants. For what output will
c+P
the expense be least?
b
dC
=a−
= 0 for maximum or minimum;
dP
(c + P )2
r
b
b
hence a =
and P = ±
− c.
2
(c + P )
a
r
b
As the output cannot be negative, P = +
− c.
a
Now
d2 C
b(2c + 2P )
=+
,
2
dP
(c + P )4
r
which is positive for all the values of P ; hence P = +
to a minimum.
b
−c corresponds
a
CALCULUS MADE EASY
114
(5) The total cost per hour C of lighting a building with N lamps
of a certain kind is
C=N
Cl EP Ce
+
t
1000
,
where E is the commercial efficiency (watts per candle),
P is the candle power of each lamp,
t is the average life of each lamp in hours,
Cl = cost of renewal in pence per hour of use,
Ce = cost of energy per 1000 watts per hour.
Moreover, the relation connecting the average life of a lamp with
the commercial efficiency at which it is run is approximately t = mE n ,
where m and n are constants depending on the kind of lamp.
Find the commercial efficiency for which the total cost of lighting
will be least.
Cl −n P Ce
E +
E ,
C=N
m
1000
dC
P Ce
nCl −(n+1)
=
−
E
=0
dE
1000
m
We have
for maximum or minimum.
E
n+1
1000 × nCl
=
mP Ce
r
and E =
n+1
1000 × nCl
.
mP Ce
This is clearly for minimum, since
d2 C
nCl −(n+2)
= (n + 1)
E
,
2
dE
m
CURVATURE OF CURVES
115
which is positive for a positive value of E.
For a particular type of 16 candle-power lamps, Cl = 17 pence,
Ce = 5 pence; and it was found that m = 10 and n = 3.6.
r
1000 × 3.6 × 17
= 2.6 watts per candle-power.
E = 4.6
10 × 16 × 5
Exercises X.
(You are advised to plot the graph of any numerical
example.) (See p. 258 for the Answers.)
(1) Find the maxima and minima of
y = x3 + x2 − 10x + 8.
dy
d2 y
b
x − cx2 , find expressions for
, and for
, also
a
dx
dx2
find the value of x which makes y a maximum or a minimum, and show
(2) Given y =
whether it is maximum or minimum.
(3) Find how many maxima and how many minima there are in the
curve, the equation to which is
y =1−
x2 x4
+ ;
2
24
and how many in that of which the equation is
y =1−
x 2 x4
x6
+
−
.
2
24 720
(4) Find the maxima and minima of
y = 2x + 1 +
5
.
x2
CALCULUS MADE EASY
116
(5) Find the maxima and minima of
y=
x2
3
.
+x+1
(6) Find the maxima and minima of
y=
5x
.
2 + x2
(7) Find the maxima and minima of
y=
3x
x
+ + 5.
−3 2
x2
(8) Divide a number N into two parts in such a way that three times
the square of one part plus twice the square of the other part shall be
a minimum.
(9) The efficiency u of an electric generator at different values of
output x is expressed by the general equation:
u=
x
;
a + bx + cx2
where a is a constant depending chiefly on the energy losses in the iron
and c a constant depending chiefly on the resistance of the copper parts.
Find an expression for that value of the output at which the efficiency
will be a maximum.
(10) Suppose it to be known that consumption of coal by a certain
steamer may be represented by the formula y = 0.3+0.001v 3 ; where y is
the number of tons of coal burned per hour and v is the speed expressed
in nautical miles per hour. The cost of wages, interest on capital, and
depreciation of that ship are together equal, per hour, to the cost of
CURVATURE OF CURVES
117
1 ton of coal. What speed will make the total cost of a voyage of 1000
nautical miles a minimum? And, if coal costs 10 shillings per ton, what
will that minimum cost of the voyage amount to?
(11) Find the maxima and minima of
y=±
xp
x(10 − x).
6
(12) Find the maxima and minima of
y = 4x3 − x2 − 2x + 1.
CHAPTER XIII.
OTHER USEFUL DODGES.
Partial Fractions.
We have seen that when we differentiate a fraction we have to perform
a rather complicated operation; and, if the fraction is not itself a simple
one, the result is bound to be a complicated expression. If we could
split the fraction into two or more simpler fractions such that their
sum is equivalent to the original fraction, we could then proceed by
differentiating each of these simpler expressions. And the result of
differentiating would be the sum of two (or more) differentials, each one
of which is relatively simple; while the final expression, though of course
it will be the same as that which could be obtained without resorting
to this dodge, is thus obtained with much less effort and appears in a
simplified form.
Let us see how to reach this result. Try first the job of adding
two fractions together to form a resultant fraction. Take, for example,
2
1
and
. Every schoolboy can add these
the two fractions
x+1
x−1
3x + 1
together and find their sum to be 2
. And in the same way he can
x −1
add together three or more fractions. Now this process can certainly
be reversed: that is to say, that if this last expression were given, it
OTHER USEFUL DODGES
119
is certain that it can somehow be split back again into its original
components or partial fractions. Only we do not know in every case
that may be presented to us how we can so split it. In order to find
this out we shall consider a simple case at first. But it is important
to bear in mind that all which follows applies only to what are called
“proper” algebraic fractions, meaning fractions like the above, which
have the numerator of a lesser degree than the denominator; that is,
those in which the highest index of x is less in the numerator than in
x2 + 2
,
the denominator. If we have to deal with such an expression as 2
x −1
3
we can simplify it by division, since it is equivalent to 1 + 2
; and
x −1
3
is a proper algebraic fraction to which the operation of splitting
2
x −1
into partial fractions can be applied, as explained hereafter.
Case I. If we perform many additions of two or more fractions
the denominators of which contain only terms in x, and no terms in
x2 , x3 , or any other powers of x, we always find that the denominator
of the final resulting fraction is the product of the denominators of
the fractions which were added to form the result. It follows that by
factorizing the denominator of this final fraction, we can find every one
of the denominators of the partial fractions of which we are in search.
3x + 1
Suppose we wish to go back from 2
to the components which
x −1
1
2
we know are
and
. If we did not know what those compox+1
x−1
nents were we can still prepare the way by writing:
3x + 1
3x + 1
=
=
+
,
x2 − 1
(x + 1)(x − 1)
x+1 x−1
leaving blank the places for the numerators until we know what to put
CALCULUS MADE EASY
120
there. We always may assume the sign between the partial fractions
to be plus, since, if it be minus, we shall simply find the corresponding
numerator to be negative. Now, since the partial fractions are proper
fractions, the numerators are mere numbers without x at all, and we
can call them A, B, C . . . as we please. So, in this case, we have:
A
B
3x + 1
=
+
.
2
x −1
x+1 x−1
If now we perform the addition of these two partial fractions, we
A(x − 1) + B(x + 1)
3x + 1
get
; and this must be equal to
.
(x + 1)(x − 1)
(x + 1)(x − 1)
And, as the denominators in these two expressions are the same, the
numerators must be equal, giving us:
3x + 1 = A(x − 1) + B(x + 1).
Now, this is an equation with two unknown quantities, and it would
seem that we need another equation before we can solve them and find
A and B. But there is another way out of this difficulty. The equation
must be true for all values of x; therefore it must be true for such values
of x as will cause x − 1 and x + 1 to become zero, that is for x = 1 and
for x = −1 respectively. If we make x = 1, we get 4 = (A×0)+(B ×2),
so that B = 2; and if we make x = −1, we get −2 = (A×−2)+(B ×0),
so that A = 1. Replacing the A and B of the partial fractions by these
1
2
new values, we find them to become
and
; and the thing is
x+1
x−1
done.
4x2 + 2x − 14
. The
As a farther example, let us take the fraction 3
x + 3x2 − x − 3
denominator becomes zero when x is given the value 1; hence x − 1 is
a factor of it, and obviously then the other factor will be x2 + 4x + 3;
OTHER USEFUL DODGES
121
and this can again be decomposed into (x + 1)(x + 3). So we may write
the fraction thus:
A
B
C
4x2 + 2x − 14
=
+
+
,
3
2
x + 3x − x − 3
x+1 x−1 x+3
making three partial factors.
Proceeding as before, we find
4x2 + 2x − 14 = A(x − 1)(x + 3) + B(x + 1)(x + 3) + C(x + 1)(x − 1).
Now, if we make x = 1, we get:
−8 = (A × 0) + B(2 × 4) + (C × 0);
that is, B = −1.
If x = −1, we get:
−12 = A(−2 × 2) + (B × 0) + (C × 0);
whence A = 3.
If x = −3, we get:
16 = (A × 0) + (B × 0) + C(−2 × −4);
whence C = 2.
So then the partial fractions are:
1
2
3
−
+
,
x+1 x−1 x+3
which is far easier to differentiate with respect to x than the complicated
expression from which it is derived.
CALCULUS MADE EASY
122
Case II. If some of the factors of the denominator contain terms
in x2 , and are not conveniently put into factors, then the corresponding
numerator may contain a term in x, as well as a simple number; and
hence it becomes necessary to represent this unknown numerator not
by the symbol A but by Ax + B; the rest of the calculation being made
as before.
−x2 − 3
.
Try, for instance:
(x2 + 1)(x + 1)
Ax + B
C
−x2 − 3
= 2
+
;
2
(x + 1)(x + 1)
x +1
x+1
−x2 − 3 = (Ax + B)(x + 1) + C(x2 + 1).
Putting x = −1, we get −4 = C × 2; and C = −2;
hence
and
−x2 − 3 = (Ax + B)(x + 1) − 2x2 − 2;
x2 − 1 = Ax(x + 1) + B(x + 1).
Putting x = 0, we get −1 = B;
hence
x2 − 1 = Ax(x + 1) − x − 1;
and
or x2 + x = Ax(x + 1);
x + 1 = A(x + 1),
so that A = 1, and the partial fractions are:
x−1
2
−
.
2
x +1 x+1
Take as another example the fraction
x3 − 2
.
(x2 + 1)(x2 + 2)
OTHER USEFUL DODGES
123
We get
Ax + B Cx + D
x3 − 2
= 2
+ 2
2
2
(x + 1)(x + 2)
x +1
x +2
2
(Ax + B)(x + 2) + (Cx + D)(x2 + 1)
=
.
(x2 + 1)(x2 + 2)
In this case the determination of A, B, C, D is not so easy. It
will be simpler to proceed as follows: Since the given fraction and
the fraction found by adding the partial fractions are equal, and have
identical denominators, the numerators must also be identically the
same. In such a case, and for such algebraical expressions as those with
which we are dealing here, the coefficients of the same powers of x are
equal and of same sign.
Hence, since
x3 − 2 = (Ax + B)(x2 + 2) + (Cx + D)(x2 + 1)
= (A + C)x3 + (B + D)x2 + (2A + C)x + 2B + D,
we have 1 = A + C; 0 = B + D (the coefficient of x2 in the left
expression being zero); 0 = 2A + C; and −2 = 2B + D. Here are
four equations, from which we readily obtain A = −1; B = −2; C = 2;
2(x + 1) x + 2
− 2
. This method
D = 0; so that the partial fractions are 2
x +2
x +1
can always be used; but the method shown first will be found the
quickest in the case of factors in x only.
Case III. When, among the factors of the denominator there are
some which are raised to some power, one must allow for the possible existence of partial fractions having for denominator the several
powers of that factor up to the highest. For instance, in splitting the
CALCULUS MADE EASY
124
3x2 − 2x + 1
we must allow for the possible existence of a
(x + 1)2 (x − 2)
denominator x + 1 as well as (x + 1)2 and (x − 2).
fraction
It maybe thought, however, that, since the numerator of the fraction
the denominator of which is (x + 1)2 may contain terms in x, we must
allow for this in writing Ax + B for its numerator, so that
Ax + B
C
D
3x2 − 2x + 1
=
+
+
.
2
2
(x + 1) (x − 2)
(x + 1)
x+1 x−2
If, however, we try to find A, B, C and D in this case, we fail, because
we get four unknowns; and we have only three relations connecting
them, yet
3x2 − 2x + 1
x−1
1
1
=
+
+
.
2
2
(x + 1) (x − 2)
(x + 1)
x+1 x−2
But if we write
3x2 − 2x + 1
A
B
C
+
=
+
,
2
2
(x + 1) (x − 2)
(x + 1)
x+1 x−2
we get
3x2 − 2x + 1 = A(x − 2) + B(x + 1)(x − 2) + C(x + 1)2 ,
which gives C = 1 for x = 2. Replacing C by its value, transposing,
gathering like terms and dividing by x − 2, we get −2x = A + B(x + 1),
which gives A = −2 for x = −1. Replacing A by its value, we get
2x = −2 + B(x + 1).
Hence B = 2; so that the partial fractions are:
2
2
1
−
+
,
2
x + 1 (x + 1)
x−2
OTHER USEFUL DODGES
125
x−1
1
1
+
stated above as being the fractions
+
2
x + 1 (x + 1)
x−2
3x2 − 2x + 1
from which
was obtained. The mystery is cleared if we
(x + 1)2 (x − 2)
x−1
1
observe that
can itself be split into the two fractions
−
2
(x + 1)
x+1
2
, so that the three fractions given are really equivalent to
(x + 1)2
instead of
1
1
2
1
2
2
1
+
−
+
=
−
+
,
2
2
x + 1 x + 1 (x + 1)
x−2
x + 1 (x + 1)
x−2
which are the partial fractions obtained.
We see that it is sufficient to allow for one numerical term in each
numerator, and that we always get the ultimate partial fractions.
When there is a power of a factor of x2 in the denominator, however,
the corresponding numerators must be of the form Ax+B; for example,
(2x2
Ax + B
Cx + D
E
3x − 1
=
+ 2
+
,
2
2
2
− 1) (x + 1)
(2x − 1)
2x − 1
x+1
which gives
3x − 1 = (Ax + B)(x + 1) + (Cx + D)(x + 1)(2x2 − 1) + E(2x2 − 1)2 .
For x = −1, this gives E = −4. Replacing, transposing, collecting
like terms, and dividing by x + 1, we get
16x3 − 16x2 + 3 = 2Cx3 + 2Dx2 + x(A − C) + (B − D).
Hence 2C = 16 and C = 8; 2D = −16 and D = −8; A − C = 0 or
A − 8 = 0 and A = 8, and finally, B − D = 3 or B = −5. So that we
obtain as the partial fractions:
(8x − 5)
8(x − 1)
4
+
−
.
2
2
2
(2x − 1)
2x − 1
x+1
CALCULUS MADE EASY
126
It is useful to check the results obtained. The simplest way is to
replace x by a single value, say +1, both in the given expression and in
the partial fractions obtained.
Whenever the denominator contains but a power of a single factor,
a very quick method is as follows:
4x + 1
, let x + 1 = z; then x = z − 1.
Taking, for example,
(x + 1)3
Replacing, we get
4(z − 1) + 1
4z − 3
4
3
=
=
−
.
z3
z3
z2 z3
The partial fractions are, therefore,
4
3
−
.
2
(x + 1)
(x + 1)3
Application to differentiation. Let it be required to differentiate
5 − 4x
y= 2
; we have
6x + 7x − 3
dy
(6x2 + 7x − 3) × 4 + (5 − 4x)(12x + 7)
=−
dx
(6x2 + 7x − 3)2
2
24x − 60x − 23
=
.
(6x2 + 7x − 3)2
If we split the given expression into
1
2
−
,
3x − 1 2x + 3
we get, however,
3
4
dy
=−
+
,
2
dx
(3x − 1)
(2x + 3)2
OTHER USEFUL DODGES
127
which is really the same result as above split into partial fractions.
But the splitting, if done after differentiating, is more complicated, as
will easily be seen. When we shall deal with the integration of such
expressions, we shall find the splitting into partial fractions a precious
auxiliary (see p. 228).
Exercises XI.
(See page 259 for Answers.)
Split into fractions:
(1)
(3)
3x + 5
.
(x − 3)(x + 4)
(2)
3x + 5
.
+ x − 12
(4)
x2
(5)
x−8
.
(2x + 3)(3x − 2)
(7)
x2 − 3x + 1
.
(x − 1)(x + 2)(x − 3)
(8)
3x − 4
.
(x − 1)(x − 2)
x2
x+1
.
− 7x + 12
(6)
x2 − 13x + 26
.
(x − 2)(x − 3)(x − 4)
5x2 + 7x + 1
.
(2x + 1)(3x − 2)(3x + 1)
(9)
x2
.
x3 − 1
(10)
x4 + 1
.
x3 + 1
(11)
5x2 + 6x + 4
.
(x + 1)(x2 + x + 1)
(12)
x
.
(x − 1)(x − 2)2
(13)
x
.
(x2 − 1)(x + 1)
(14)
x+3
.
(x + 2)2 (x − 1)
CALCULUS MADE EASY
(15)
3x2 + 2x + 1
.
(x + 2)(x2 + x + 1)2
(16)
5x2 + 8x − 12
.
(x + 4)3
(17)
7x2 + 9x − 1
.
(3x − 2)4
(18)
x2
.
(x3 − 8)(x − 2)
128
Differential of an Inverse Function.
Consider the function (see p. 13) y = 3x; it can be expressed in the
y
form x = ; this latter form is called the inverse function to the one
3
originally given.
dy
y dx
1
If y = 3x,
= 3; if x = ,
= , and we see that
dx
3 dy
3
1
dy
=
dx
dx
dy
Consider y = 4x2 ,
Here again
dy dx
×
= 1.
dx dy
dy
= 8x; the inverse function is
dx
1
y2
x=
,
2
or
and
dx
1
1
1
= √ =
=
.
dy
4 y
4 × 2x
8x
dy dx
×
= 1.
dx dy
It can be shown that for all functions which can be put into the
inverse form, one can always write
dy dx
×
= 1 or
dx dy
dy
1
=
.
dx
dx
dy
129
OTHER USEFUL DODGES
It follows that, being given a function, if it be easier to differentiate
the inverse function, this may be done, and the reciprocal of the differential coefficient of the inverse function gives the differential coefficient
of the given function itself.
r
As an example, suppose that we wish to differentiate y =
2
3
− 1.
x
3
We have seen one way of doing this, by writing u = − 1, and finding
x
du
dy
and
. This gives
du
dx
3
dy
r
=−
.
dx
3
2x2
−1
x
If we had forgotten how to proceed by this method, or wished to
check our result by some other way of obtaining the differential coefficient, or for any other reason we could not use the ordinary method,
3
.
we can proceed as follows: The inverse function is x =
1 + y2
dx
3 × 2y
6y
=−
=−
;
2
2
dy
(1 + y )
(1 + y 2 )2
hence
2
3
1+ −1
dy
(1 + y 2 )2
1
3
x
r
r
=−
=
=−
=−
.
dx
dx
6y
3
3
6× 2 −1
2x2
−1
dy
x
x
1
Let us take as an other example y = √
.
3
θ+5
1
The inverse function is θ = 3 − 5 or θ = y −3 − 5, and
y
p
dθ
= −3y −4 = −3 3 (θ + 5)4 .
dy
CALCULUS MADE EASY
It follows that
130
1
dy
=− p
, as might have been found othdx
3 (θ + 5)4
erwise.
We shall find this dodge most useful later on; meanwhile you are
advised to become familiar with it by verifying by its means the results
obtained in Exercises I. (p. 24), Nos. 5, 6, 7; Examples (p. 67), Nos.
1, 2, 4; and Exercises VI. (p. 72), Nos. 1, 2, 3 and 4.
You will surely realize from this chapter and the preceding, that in
many respects the calculus is an art rather than a science: an art only
to be acquired, as all other arts are, by practice. Hence you should
work many examples, and set yourself other examples, to see if you can
work them out, until the various artifices become familiar by use.
CHAPTER XIV.
ON TRUE COMPOUND INTEREST AND THE LAW
OF ORGANIC GROWTH.
Let there be a quantity growing in such a way that the increment of
its growth, during a given time, shall always be proportional to its own
magnitude. This resembles the process of reckoning interest on money
at some fixed rate; for the bigger the capital, the bigger the amount of
interest on it in a given time.
Now we must distinguish clearly between two cases, in our calculation, according as the calculation is made by what the arithmetic books
call “simple interest,” or by what they call “compound interest.” For
in the former case the capital remains fixed, while in the latter the
interest is added to the capital, which therefore increases by successive
additions.
(1) At simple interest. Consider a concrete case. Let the capital at
start be £100, and let the rate of interest be 10 per cent. per annum.
Then the increment to the owner of the capital will be £10 every year.
Let him go on drawing his interest every year, and hoard it by putting
it by in a stocking, or locking it up in his safe. Then, if he goes on for
10 years, by the end of that time he will have received 10 increments
132
CALCULUS MADE EASY
of £10 each, or £100, making, with the original £100, a total of £200
in all. His property will have doubled itself in 10 years. If the rate of
interest had been 5 per cent., he would have had to hoard for 20 years
to double his property. If it had been only 2 per cent., he would have
had to hoard for 50 years. It is easy to see that if the value of the yearly
1
interest is of the capital, he must go on hoarding for n years in order
n
to double his property.
y
Or, if y be the original capital, and the yearly interest is , then,
n
at the end of n years, his property will be
y+n
y
= 2y.
n
(2) At compound interest. As before, let the owner begin with a
capital of £100, earning interest at the rate of 10 per cent. per annum;
but, instead of hoarding the interest, let it be added to the capital
each year, so that the capital grows year by year. Then, at the end
of one year, the capital will have grown to £110; and in the second
year (still at 10%) this will earn £11 interest. He will start the third
year with £121, and the interest on that will be £12. 2s.; so that he
starts the fourth year with £133. 2s., and so on. It is easy to work
it out, and find that at the end of the ten years the total capital will
have grown to £259. 7s. 6d. In fact, we see that at the end of each
year, each pound will have earned
1
10
of a pound, and therefore, if
this is always added on, each year multiplies the capital by
11
;
10
and if
continued for ten years (which will multiply by this factor ten times
over) will multiply the original capital by 2.59374. Let us put this into
1
symbols. Put y0 for the original capital; for the fraction added on at
n
ON TRUE COMPOUND INTEREST
133
each of the n operations; and yn for the value of the capital at the end
of the nth operation. Then
n
1
yn = y0 1 +
.
n
But this mode of reckoning compound interest once a year, is really
not quite fair; for even during the first year the £100 ought to have been
growing. At the end of half a year it ought to have been at least £105,
and it certainly would have been fairer had the interest for the second
half of the year been calculated on £105. This would be equivalent to
calling it 5% per half-year; with 20 operations, therefore, at each of
which the capital is multiplied by
21
.
20
If reckoned this way, by the end
of ten years the capital would have grown to £265. 6s. 7d.; for
(1 +
1 20
)
20
= 2.653.
But, even so, the process is still not quite fair; for, by the end of
the first month, there will be some interest earned; and a half-yearly
reckoning assumes that the capital remains stationary for six months at
a time. Suppose we divided the year into 10 parts, and reckon a one-percent. interest for each tenth of the year. We now have 100 operations
lasting over the ten years; or
yn = £100 1 +
1 100
100
;
which works out to £270. 9s. 7 12 d.
Even this is not final. Let the ten years be divided into 1000 periods,
each of
1
100
of a year; the interest being
1
10
per cent. for each such period;
then
yn = £100 1 +
1000
1
1000
;
CALCULUS MADE EASY
134
which works out to £271. 13s. 10d.
Go even more minutely, and divide the ten years into 10, 000 parts,
each
1
1000
1
of a year, with interest at 100
of 1 per cent. Then
10,000
1
yn = £100 1 + 10,000
;
which amounts to £271. 16s. 3 21 d.
Finally, it will be seen that what we
are trying
n to find is in reality
1
the ultimate value of the expression 1 +
, which, as we see, is
n
greater than 2; and which, as we take n larger and larger, grows closer
and closer to a particular limiting value. However big you make n, the
value of this expression grows nearer and nearer to the figure
2.71828 . . .
a number never to be forgotten.
Let us take geometrical illustrations of these things. In Fig. 36,
OP stands for the original value. OT is the whole time during which
the value is growing. It is divided into 10 periods, in each of which
dy
there is an equal step up. Here
is a constant; and if each step up
dx
1
is 10 of the original OP , then, by 10 such steps, the height is doubled.
If we had taken 20 steps, each of half the height shown, at the end the
1
height would still be just doubled. Or n such steps, each of
of the
n
original height OP , would suffice to double the height. This is the case
of simple interest. Here is 1 growing till it becomes 2.
In Fig. 37, we have the corresponding illustration of the geometrical
1
progression. Each of the successive ordinates is to be 1 + , that is,
n
n+1
times as high as its predecessor. The steps up are not equal,
n
ON TRUE COMPOUND INTEREST
135
U
P
2
1
O
1
2
3
4
5
6
7
8
9
T
Fig. 36.
1
of the ordinate at that part of the curve.
n
1
If we had literally 10 steps, with 1 + 10
for the multiplying factor,
because each step up is now
1 10
)
10
or 2.594 times the original 1. But if
1
only we take n sufficiently large (and the corresponding sufficiently
n
n
1
small), then the final value 1 +
to which unity will grow will
n
be 2.71828.
the final total would be (1 +
U
2.7182
P
1
O
1
2
3
4
5
6
7
8
9
T
Fig. 37.
Epsilon. To this mysterious number 2.7182818 etc., the mathemati-
CALCULUS MADE EASY
136
cians have assigned as a symbol the Greek letter (pronounced epsilon). All schoolboys know that the Greek letter π (called pi ) stands
for 3.141592 etc.; but how many of them know that epsilon means
2.71828? Yet it is an even more important number than π!
What, then, is epsilon?
Suppose we were to let 1 grow at simple interest till it became 2;
then, if at the same nominal rate of interest, and for the same time,
we were to let 1 grow at true compound interest, instead of simple, it
would grow to the value epsilon.
This process of growing proportionately, at every instant, to the
magnitude at that instant, some people call a logarithmic rate of growing. Unit logarithmic rate of growth is that rate which in unit time will
cause 1 to grow to 2.718281. It might also be called the organic rate
of growing: because it is characteristic of organic growth (in certain
circumstances) that the increment of the organism in a given time is
proportional to the magnitude of the organism itself.
If we take 100 per cent. as the unit of rate, and any fixed period
as the unit of time, then the result of letting 1 grow arithmetically at
unit rate, for unit time, will be 2, while the result of letting 1 grow
logarithmically at unit rate, for the same time, will be 2.71828 . . . .
A little more about Epsilon. We have seen
nwe require to know
that
1
, when n becomes
what value is reached by the expression 1 +
n
indefinitely great. Arithmetically, here are tabulated a lot of values
(which anybody can calculate out by the help of an ordinary table of
logarithms) got by assuming n = 2; n = 5; n = 10; and so on, up to
THE LAW OF ORGANIC GROWTH
137
n = 10, 000.
(1 + 12 )2
= 2.25.
(1 + 15 )5
= 2.488.
(1 +
1 10
)
10
= 2.594.
(1 +
1 20
)
20
= 2.653.
(1 +
1 100
)
100
= 2.705.
(1 +
1
)1000
1000
= 2.7169.
(1 +
1
)10,000
10,000
= 2.7181.
It is, however, worth while to find another way of calculating this
immensely important figure.
Accordingly, we will avail ourselves of the binomial theorem, and
n
1
expand the expression 1 +
in that well-known way.
n
The binomial theorem gives the rule that
an−2 b2
an−1 b
+ n(n − 1)
1!
2!
n−3 3
a b
+ n(n − 1)(n − 2)
+ etc.
3!
(a + b)n = an + n
1
, we get
n
n
1 n−1
1 (n − 1)(n − 2)
1
1+
=1+1+
+
n
2!
n
3!
n2
1 (n − 1)(n − 2)(n − 3)
+
+ etc.
4!
n3
Putting a = 1 and b =
CALCULUS MADE EASY
138
Now, if we suppose n to become indefinitely great, say a billion, or
a billion billions, then n − 1, n − 2, and n − 3, etc., will all be sensibly
equal to n; and then the series becomes
=1+1+
1
1
1
+ + + etc. . . .
2! 3! 4!
By taking this rapidly convergent series to as many terms as we
please, we can work out the sum to any desired point of accuracy. Here
is the working for ten terms:
1.000000
dividing by 1
1.000000
dividing by 2
0.500000
dividing by 3
0.166667
dividing by 4
0.041667
dividing by 5
0.008333
dividing by 6
0.001389
dividing by 7
0.000198
dividing by 8
0.000025
dividing by 9
0.000002
Total 2.718281
is incommensurable with 1, and resembles π in being an interminable non-recurrent decimal.
The Exponential Series. We shall have need of yet another series.
Let us,
use of the binomial theorem, expand the ex againmaking
nx
1
pression 1 +
, which is the same as x when we make n indefin
THE LAW OF ORGANIC GROWTH
139
nitely great.
2
1
1
nx−2
1
1
n
n
+ nx(nx − 1)
x = 1nx + nx
1!
2!
3
1
1nx−3
n
+ nx(nx − 1)(nx − 2)
+ etc.
3!
1 n2 x2 − nx
1 n3 x3 − 3n2 x2 + 2nx
=1+x+ ·
+
·
+ etc.
2!
n2
3!
n3
x
3x2 2x
x2 −
x3 −
+ 2
n +
n
n + etc.
=1+x+
2!
3!
nx−1
But, when n is made indefinitely great, this simplifies down to the
following:
x2 x 3 x4
+
+
+ etc. . . .
2!
3!
4!
This series is called the exponential series.
x = 1 + x +
The great reason why is regarded of importance is that x possesses
a property, not possessed by any other function of x, that when you
differentiate it its value remains unchanged ; or, in other words, its
differential coefficient is the same as itself. This can be instantly seen
by differentiating it with respect to x, thus:
or
d(x )
2x
3x2
4x3
=0+1+
+
+
dx
1·2 1·2·3 1·2·3·4
5x4
+
+ etc.
1·2·3·4·5
x2
x3
x4
=1+x+
+
+
+ etc.,
1·2 1·2·3 1·2·3·4
which is exactly the same as the original series.
CALCULUS MADE EASY
140
Now we might have gone to work the other way, and said: Go to;
let us find a function of x, such that its differential coefficient is the
same as itself. Or, is there any expression, involving only powers of x,
which is unchanged by differentiation? Accordingly; let us assume as
a general expression that
y = A + Bx + Cx2 + Dx3 + Ex4 + etc.,
(in which the coefficients A, B, C, etc. will have to be determined),
and differentiate it.
dy
= B + 2Cx + 3Dx2 + 4Ex3 + etc.
dx
Now, if this new expression is really to be the same as that from
B
A
which it was derived, it is clear that A must = B; that C =
=
;
2
1·2
A
D
A
C
=
; that E =
=
, etc.
that D =
3
1·2·3
4
1·2·3·4
The law of change is therefore that
x
x2
x3
x4
y =A 1+ +
+
+
+ etc. .
1 1·2 1·2·3 1·2·3·4
If, now, we take A = 1 for the sake of further simplicity, we have
y =1+
x
x2
x3
x4
+
+
+
+ etc.
1 1·2 1·2·3 1·2·3·4
Differentiating it any number of times will give always the same
series over again.
If, now, we take the particular case of A = 1, and evaluate the
THE LAW OF ORGANIC GROWTH
141
series, we shall get simply
when x = 1,
y = 2.718281 etc.;
that is, y = ;
when x = 2,
y = (2.718281 etc.)2 ;
that is, y = 2 ;
when x = 3,
y = (2.718281 etc.)3 ;
that is, y = 3 ;
and therefore
when x = x,
y = (2.718281 etc.)x ;
that is, y = x ,
thus finally demonstrating that
x = 1 +
x2
x3
x4
x
+
+
+
+ etc.
1 1·2 1·2·3 1·2·3·4
[Note.—How to read exponentials. For the benefit of those who
have no tutor at hand it may be of use to state that x is read as
“epsilon to the eksth power ;” or some people read it “exponential eks.”
So pt is read “epsilon to the pee-teeth-power ” or “exponential pee tee.”
Take some similar expressions:—Thus, −2 is read “epsilon to the minus
two power ” or “exponential minus two.” −ax is read “epsilon to the
minus ay-eksth” or “exponential minus ay-eks.”]
Of course it follows that y remains unchanged if differentiated with
respect to y. Also ax , which is equal to (a )x , will, when differentiated
with respect to x, be aax , because a is a constant.
Natural or Naperian Logarithms.
Another reason why is important is because it was made by Napier,
the inventor of logarithms, the basis of his system. If y is the value of
142
CALCULUS MADE EASY
x , then x is the logarithm, to the base , of y. Or, if
y = x ,
then
x = log y.
The two curves plotted in Figs. 38 and 39 represent these equations.
O
1
1.5
2
1
1.65
2.71
4.50
7.39
1
2
3
4
8
0
0.69
1.10
1.39
2.08
1
1.5
Fig. 38.
2
x
O
ǫ
log
=
y
0.5
=
y
8
7
6
5
4
3
2
1
y
0.5
ǫx
y
8
7
6
5
4
3
2
1
0
x
The points calculated are:

 x
For Fig. 38
 y

 y
For Fig. 39
 x
1
2
x
Fig. 39.
It will be seen that, though the calculations yield different points
for plotting, yet the result is identical. The two equations really mean
the same thing.
THE LAW OF ORGANIC GROWTH
143
As many persons who use ordinary logarithms, which are calculated to base 10 instead of base , are unfamiliar with the “natural”
logarithms, it may be worth while to say a word about them. The ordinary rule that adding logarithms gives the logarithm of the product
still holds good; or
log a + log b = log ab.
Also the rule of powers holds good;
n × log a = log an .
But as 10 is no longer the basis, one cannot multiply by 100 or 1000
by merely adding 2 or 3 to the index. One can change the natural
logarithm to the ordinary logarithm simply by multiplying it by 0.4343;
or
log10 x = 0.4343 × log x,
and conversely,
log x = 2.3026 × log10 x.
Exponential and Logarithmic Equations.
Now let us try our hands at differentiating certain expressions that
contain logarithms or exponentials.
Take the equation:
y = log x.
First transform this into
y = x,
whence, since the differential of y with regard to y is the original
function unchanged (see p. 139),
dx
= y ,
dy
144
CALCULUS MADE EASY
A Useful Table of “Naperian Logarithms”
(Also called Natural Logarithms or Hyperbolic Logarithms)
Number
log
Number
log
1
0.0000
6
1.7918
1.1
0.0953
7
1.9459
1.2
0.1823
8
2.0794
1.5
0.4055
9
2.1972
1.7
0.5306
10
2.3026
2.0
0.6931
20
2.9957
2.2
0.7885
50
3.9120
2.5
0.9163
100
4.6052
2.7
0.9933
200
5.2983
2.8
1.0296
500
6.2146
3.0
1.0986
1, 000
6.9078
3.5
1.2528
2, 000
7.6009
4.0
1.3863
5, 000
8.5172
4.5
1.5041
10, 000
9.2103
5.0
1.6094
20, 000
9.9035
THE LAW OF ORGANIC GROWTH
145
and, reverting from the inverse to the original function,
1
1
1
dy
=
= y = .
dx
dx
x
dy
Now this is a very curious result. It may be written
d(log x)
= x−1 .
dx
Note that x−1 is a result that we could never have got by the rule for
differentiating powers. That rule (page 24) is to multiply by the power,
and reduce the power by 1. Thus, differentiating x3 gave us 3x2 ; and
differentiating x2 gave 2x1 . But differentiating x0 does not give us
x−1 or 0 × x−1 , because x0 is itself = 1, and is a constant. We shall
have to come back to this curious fact that differentiating log x gives
1
us when we reach the chapter on integrating.
x
Now, try to differentiate
y = log (x + a),
y = x + a;
that is
we have
d(x + a)
= y , since the differential of y remains y .
dy
dx
= y = x + a;
dy
hence, reverting to the original function (see p. 128), we get
dy
1
1
=
=
.
dx
dx
x+a
dy
This gives
CALCULUS MADE EASY
Next try
146
y = log10 x.
First change to natural logarithms by multiplying by the modulus
0.4343. This gives us
y = 0.4343 log x;
0.4343
dy
=
.
dx
x
whence
The next thing is not quite so simple. Try this:
y = ax .
Taking the logarithm of both sides, we get
log y = x log a,
or
x=
Since
log y
1
=
× log y.
log a
log a
1
is a constant, we get
log a
dx
1
1
1
=
× = x
;
dy
log a y
a × log a
hence, reverting to the original function.
dy
1
=
= ax × log a.
dx
dx
dy
THE LAW OF ORGANIC GROWTH
147
We see that, since
dx
1
1
= ×
,
dy
y log a
dx dy
×
= 1 and
dy dx
1 dy
×
= log a.
y dx
We shall find that whenever we have an expression such as log y =
1 dy
= the differential coefficient of
a function of x, we always have
y dx
the function of x, so that we could have written at once, from log y =
x log a,
1 dy
= log a and
y dx
dy
= ax log a.
dx
Let us now attempt further examples.
Examples.
(1) y = −ax . Let −ax = z; then y = z .
dy
= z ;
dx
dz
= −a;
dx
hence
dy
= −a−ax .
dx
Or thus:
log y = −ax;
x2
(2) y = 3 . Let
1 dy
= −a;
y dx
dy
= −ay = −a−ax .
dx
x2
= z; then y = z .
3
dy
= z ;
dz
dz
2x
=
;
dx
3
dy
2x x2
=
3.
dx
3
Or thus:
x2
log y = ;
3
1 dy
2x
=
;
y dx
3
dy
2x x2
=
3.
dx
3
CALCULUS MADE EASY
148
2x
(3) y = x+1 .
2x
1 dy
2(x + 1) − 2x
,
=
;
x+1
y dx
(x + 1)2
2x
2
dy
x+1 .
=
dx
(x + 1)2
log y =
hence
2x
Check by writing
= z.
x+1
√
1
2
(4) y = x +a . log y = (x2 + a) 2 .
1 dy
x
=
1
y dx
(x2 + a) 2
and
1
√
2
x × x +a
dy
=
1 .
dx
(x2 + a) 2
1
For if (x2 + a) 2 = u and x2 + a = v, u = v 2 ,
du
1
dv
= 1;
= 2x;
dv
dx
2v 2
√
Check by writing x2 + a = z.
du
x
=
1 .
dx
(x2 + a) 2
(5) y = log(a + x3 ). Let (a + x3 ) = z; then y = log z.
dz
dy
3x2
.
= 3x2 ; hence
=
dx
dx
a + x3
√
√
(6) y = log {3x2 + a + x2 }. Let 3x2 + a + x2 = z; then y =
dy
1
= ;
dz
z
log z.
dy
1
= ;
dz
z
dz
x
= 6x + √
;
dx
x2 + a
x
√
6x + √
2
dy
x(1 + 6 x2 + a)
x
+
a
√
√
√
=
=
.
dx
3x2 + a + x2
(3x2 + x2 + a) x2 + a
THE LAW OF ORGANIC GROWTH
√
(7) y = (x + 3)2 x − 2.
log y = 2 log (x + 3) + 12 log (x − 2).
1 dy
2
1
=
+
;
y dx
(x + 3) 2(x − 2)
√
2
1
dy
2
= (x + 3) x − 2
+
.
dx
x + 3 2(x − 2)
2
(8) y = (x2 + 3)3 (x3 − 2) 3 .
log y = 3 log (x2 + 3) + 23 log (x3 − 2);
2x
2 3x2
6x
2x2
1 dy
=3 2
+
=
+
.
y dx
(x + 3) 3 x3 − 2
x2 + 3 x3 − 2
For if y = log (x2 + 3), let x2 + 3 = z and u = log z.
du
1
= ;
dz
z
dz
= 2x;
dx
du
2x
= 2
.
dx
x +3
dv
3x2
= 3
and
dx
x −2
2
dy
6x
2x2
2
3 3
3
= (x + 3) (x − 2)
+
.
dx
x2 + 3 x3 − 2
√
2 2
x +a
√
(9) y = 3
.
x3 − a
Similarly, if v = log (x3 − 2),
1
1
log (x2 + a) − log (x3 − a).
2
3
1 dy
1 2x
1 3x2
x
x2
=
−
=
−
y dx
2 x2 + a 3 x3 − a
x2 + a x3 − a
√
2 2
dy
x +a
x
x2
= √
−
.
3 3
dx
x − a x2 + a x3 − a
log y =
and
149
CALCULUS MADE EASY
(10) y =
150
1
log x
dy
=
dx
log2
1
x =−
dz
1
= ;
dx
x
dy
1
.
= p
3
dx
3x log2 x
log x × 0 − 1 ×
1
.
x log2 x
x
p
1
1
(11) y = 3 log x = (log x) 3 . Let z = log x; y = z 3 .
dy
1 2
= z− 3 ;
dz
3
ax
1
(12) y =
.
ax
log y = ax(log 1 − log ax ) = −ax log ax .
1 dy
= −ax × ax log a − a log ax .
y dx
ax
dy
1
and
=−
(x × ax+1 log a + a log ax ).
x
dx
a
Try now the following exercises.
Exercises XII.
(See page 260 for Answers.)
(1) Differentiate y = b(ax − −ax ).
(2) Find the differential coefficient with respect to t of the expression
u = at2 + 2 log t.
d(log y)
.
dt
1 abx
(4) Show that if y = ·
,
b log a
dw
(5) If w = pv n , find
.
dv
Differentiate
(3) If y = nt , find
dy
= abx .
dx
THE LAW OF ORGANIC GROWTH
151
x
(6) y = log xn .
(7) y = 3− x−1 .
(8) y = (3x2 + 1)−5x .
(9) y = log (xa + a).
√
(10) y = (3x2 − 1)( x + 1).
(11) y =
log (x + 3)
.
x+3
(12) y = ax × xa .
(13) It was shown by Lord Kelvin that the speed of signalling
through a submarine cable depends on the value of the ratio of the
external diameter of the core to the diameter of the enclosed copper
wire. If this ratio is called y, then the number of signals s that can be
sent per minute can be expressed by the formula
1
s = ay 2 log ;
y
where a is a constant depending on the length and the quality of the
materials. Show that if these are given, s will be a maximum if y =
√
1 ÷ .
(14) Find the maximum or minimum of
y = x3 − log x.
(15) Differentiate y = log (axx ).
(16) Differentiate y = (log ax)3 .
152
CALCULUS MADE EASY
The Logarithmic Curve.
Let us return to the curve which has its successive ordinates in
geometrical progression, such as that represented by the equation y =
bpx .
We can see, by putting x = 0, that b is the initial height of y.
Then when
x = 1,
y = bp;
x = 2,
y = bp2 ;
y = bp3 ,
x = 3,
etc.
Also, we see that p is the numerical value of the ratio between the
height of any ordinate and that of the next preceding it. In Fig. 40, we
have taken p as 56 ; each ordinate being
6
5
as high as the preceding one.
log y
Y
b
O
log b
1
2
3
4
Fig. 40.
5
6
X
O
1
2
3
4
5
6
X
Fig. 41.
If two successive ordinates are related together thus in a constant
ratio, their logarithms will have a constant difference; so that, if we
should plot out a new curve, Fig. 41, with values of log y as ordinates,
it would be a straight line sloping up by equal steps. In fact, it follows
THE LAW OF ORGANIC GROWTH
153
from the equation, that
log y = log b + x · log p,
whence
log y − log b = x · log p.
Now, since log p is a mere number, and may be written as log p = a,
it follows that
y
= ax,
b
and the equation takes the new form
log
y = bax .
The Die-away Curve.
If we were to take p as a proper fraction (less than unity), the curve
would obviously tend to sink downwards, as in Fig. 42, where each
successive ordinate is
3
4
of the height of the preceding one.
The equation is still
y = bpx ;
but since p is less than one, log p will be a negative quantity, and may
be written −a; so that p = −a , and now our equation for the curve
takes the form
y = b−ax .
The importance of this expression is that, in the case where the
independent variable is time, the equation represents the course of a
154
CALCULUS MADE EASY
Y
b
O
1
2
3
4
5
6
X
Fig. 42.
great many physical processes in which something is gradually dying
away. Thus, the cooling of a hot body is represented (in Newton’s
celebrated “law of cooling”) by the equation
θt = θ0 −at ;
where θ0 is the original excess of temperature of a hot body over that
of its surroundings, θt the excess of temperature at the end of time t,
and a is a constant—namely, the constant of decrement, depending on
the amount of surface exposed by the body, and on its coefficients of
conductivity and emissivity, etc.
A similar formula,
Qt = Q0 −at ,
is used to express the charge of an electrified body, originally having a
charge Q0 , which is leaking away with a constant of decrement a; which
constant depends in this case on the capacity of the body and on the
resistance of the leakage-path.
Oscillations given to a flexible spring die out after a time; and the
dying-out of the amplitude of the motion may be expressed in a similar
THE LAW OF ORGANIC GROWTH
155
way.
In fact −at serves as a die-away factor for all those phenomena in
which the rate of decrease is proportional to the magnitude of that
dy
is proportional
which is decreasing; or where, in our usual symbols,
dt
at every moment to the value that y has at that moment. For we have
only to inspect the curve, Fig. 42 above, to see that, at every part of it,
dy
is proportional to the height y; the curve becoming flatter
the slope
dx
as y grows smaller. In symbols, thus
y = b−ax
log y = log b − ax log = log b − ax,
or
1 dy
= −a;
y dx
and, differentiating,
hence
dy
= b−ax × (−a) = −ay;
dx
or, in words, the slope of the curve is downward, and proportional to y
and to the constant a.
We should have got the same result if we had taken the equation in
the form
y = bpx ;
for then
But
giving us
as before.
dy
= bpx × log p.
dx
log p = −a;
dy
= y × (−a) = −ay,
dx
CALCULUS MADE EASY
156
The Time-constant. In the expression for the “die-away factor” −at ,
the quantity a is the reciprocal of another quantity known as “the timeconstant,” which we may denote by the symbol T . Then the die-away
t
factor will be written
−T ; and it will be seen, by making t = T that the
1
meaning of T or of
is that this is the length of time which it takes
a
for the original quantity (called θ0 or Q0 in the preceding instances) to
1
die away th part—that is to 0.3678—of its original value.
The values of x and −x are continually required in different
branches of physics, and as they are given in very few sets of mathematical tables, some of the values are tabulated on p. 157 for convenience.
As an example of the use of this table, suppose there is a hot body
cooling, and that at the beginning of the experiment (i.e. when t = 0)
it is 72◦ hotter than the surrounding objects, and if the time-constant
of its cooling is 20 minutes (that is, if it takes 20 minutes for its excess
1
of temperature to fall to part of 72◦ ), then we can calculate to what
it will have fallen in any given time t. For instance, let t be 60 minutes.
t
Then
= 60 ÷ 20 = 3, and we shall have to find the value of −3 ,
T
and then multiply the original 72◦ by this. The table shows that −3
is 0.0498. So that at the end of 60 minutes the excess of temperature
will have fallen to 72◦ × 0.0498 = 3.586◦ .
157
THE LAW OF ORGANIC GROWTH
x
x
−x
1 − −x
0.00
1.0000
1.0000
0.0000
0.10
1.1052
0.9048
0.0952
0.20
1.2214
0.8187
0.1813
0.50
1.6487
0.6065
0.3935
0.75
2.1170
0.4724
0.5276
0.90
2.4596
0.4066
0.5934
1.00
2.7183
0.3679
0.6321
1.10
3.0042
0.3329
0.6671
1.20
3.3201
0.3012
0.6988
1.25
3.4903
0.2865
0.7135
1.50
4.4817
0.2231
0.7769
1.75
5.755
0.1738
0.8262
2.00
7.389
0.1353
0.8647
2.50
12.182
0.0821
0.9179
3.00
20.086
0.0498
0.9502
3.50
33.115
0.0302
0.9698
4.00
54.598
0.0183
0.9817
4.50
90.017
0.0111
0.9889
5.00
148.41
0.0067
0.9933
5.50
244.69
0.0041
0.9959
6.00
403.43
0.00248
0.99752
7.50
1808.04
0.00055
0.99947
0.000045
0.999955
10.00
22026.5
CALCULUS MADE EASY
158
Further Examples.
(1) The strength of an electric current in a conductor at a time
t secs. after the application of the
electromotive
force producing it is
o
En
− Rt
1− L .
given by the expression C =
R
L
The time constant is .
R
Rt
If E = 10, R = 1, L = 0.01; then when t is very large the term − L
E
becomes 1, and C =
= 10; also
R
L
= T = 0.01.
R
Its value at any time may be written:
t
C = 10 − 10− 0.01 ,
the time-constant being 0.01. This means that it takes 0.01 sec. for the
1
0
variable term to fall by = 0.3678 of its initial value 10− 0.01 = 10.
t
To find the value of the current when t = 0.001 sec., say,
= 0.1,
T
−0.1 = 0.9048 (from table).
It follows that, after 0.001 sec., the variable term is 0.9048 × 10 =
9.048, and the actual current is 10 − 9.048 = 0.952.
Similarly, at the end of 0.1 sec.,
t
= 10;
T
−10 = 0.000045;
the variable term is 10 × 0.000045 = 0.00045, the current being 9.9995.
(2) The intensity I of a beam of light which has passed through a
thickness l cm. of some transparent medium is I = I0 −Kl , where I0 is
the initial intensity of the beam and K is a “constant of absorption.”
THE LAW OF ORGANIC GROWTH
159
This constant is usually found by experiments. If it be found, for
instance, that a beam of light has its intensity diminished by 18% in
passing through 10 cms. of a certain transparent medium, this means
that 82 = 100 × −K×10 or −10K = 0.82, and from the table one sees
that 10K = 0.20 very nearly; hence K = 0.02.
To find the thickness that will reduce the intensity to half its value,
one must find the value of l which satisfies the equality 50 = 100×−0.02l ,
or 0.5 = −0.02l . It is found by putting this equation in its logarithmic
form, namely,
log 0.5 = −0.02 × l × log ,
which gives
l=
−0.3010
= 34.7 centimetres nearly.
−0.02 × 0.4343
(3) The quantity Q of a radio-active substance which has not yet undergone transformation is known to be related to the initial quantity Q0
of the substance by the relation Q = Q0 −λt , where λ is a constant and
t the time in seconds elapsed since the transformation began.
For “Radium A,” if time is expressed in seconds, experiment shows
that λ = 3.85 × 10−3 . Find the time required for transforming half the
substance. (This time is called the “mean life” of the substance.)
We have 0.5 = −0.00385t .
log 0.5 = −0.00385t × log ;
and
t = 3 minutes very nearly.
160
CALCULUS MADE EASY
Exercises XIII.
(See page 260 for Answers.)
t
(1) Draw the curve y = b− T ; where b = 12, T = 8, and t is given
various values from 0 to 20.
(2) If a hot body cools so that in 24 minutes its excess of temperature
has fallen to half the initial amount, deduce the time-constant, and find
how long it will be in cooling down to 1 per cent. of the original excess.
(3) Plot the curve y = 100(1 − −2t ).
(4) The following equations give very similar curves:
(i) y =
ax
;
x+b
x
(ii) y = a(1 − − b );
x
a
(iii) y = ◦ arc tan
.
90
b
Draw all three curves, taking a = 100 millimetres; b = 30 millimetres.
(5) Find the differential coefficient of y with respect to x, if
(a) y = xx ;
(b) y = (x )x ;
x
(c) y = x .
(6) For “Thorium A,” the value of λ is 5; find the “mean life,” that
is, the time taken by the transformation of a quantity Q of “Thorium A”
equal to half the initial quantity Q0 in the expression
Q = Q0 −λt ;
t being in seconds.
THE LAW OF ORGANIC GROWTH
161
(7) A condenser of capacity K = 4 × 10−6 , charged to a potential
V0 = 20, is discharging through a resistance of 10, 000 ohms. Find the
potential V after (a) 0.1 second; (b) 0.01 second; assuming that the fall
t
of potential follows the rule V = V0 − KR .
(8) The charge Q of an electrified insulated metal sphere is reduced
from 20 to 16 units in 10 minutes. Find the coefficient µ of leakage,
if Q = Q0 × −µt ; Q0 being the initial charge and t being in seconds.
Hence find the time taken by half the charge to leak away.
(9) The damping on a telephone line can be ascertained from the relation i = i0 −βl , where i is the strength, after t seconds, of a telephonic
current of initial strength i0 ; l is the length of the line in kilometres, and
β is a constant. For the Franco-English submarine cable laid in 1910,
β = 0.0114. Find the damping at the end of the cable (40 kilometres),
and the length along which i is still 8% of the original current (limiting
value of very good audition).
(10) The pressure p of the atmosphere at an altitude h kilometres is
given by p = p0 −kh ; p0 being the pressure at sea-level (760 millimetres).
The pressures at 10, 20 and 50 kilometres being 199.2, 42.2, 0.32
respectively, find k in each case. Using the mean value of k, find the
percentage error in each case.
(11) Find the minimum or maximum of y = xx .
1
(12) Find the minimum or maximum of y = x x .
1
(13) Find the minimum or maximum of y = xa x .
CHAPTER XV.
HOW TO DEAL WITH SINES AND COSINES.
Greek letters being usual to denote angles, we will take as the usual
letter for any variable angle the letter θ (“theta”).
Let us consider the function
y = sin θ.
dθ
y
θ
O
Fig. 43.
d(sin θ)
; or, in other
dθ
words, if the angle θ varies, we have to find the relation between the
What we have to investigate is the value of
increment of the sine and the increment of the angle, both increments
being indefinitely small in themselves. Examine Fig. 43, wherein, if the
SINES AND COSINES
163
radius of the circle is unity, the height of y is the sine, and θ is the
angle. Now, if θ is supposed to increase by the addition to it of the
small angle dθ—an element of angle—the height of y, the sine, will be
increased by a small element dy. The new height y + dy will be the sine
of the new angle θ + dθ, or, stating it as an equation,
y + dy = sin(θ + dθ);
and subtracting from this the first equation gives
dy = sin(θ + dθ) − sin θ.
The quantity on the right-hand side is the difference between two
sines, and books on trigonometry tell us how to work this out. For they
tell us that if M and N are two different angles,
sin M − sin N = 2 cos
M −N
M +N
· sin
.
2
2
If, then, we put M = θ + dθ for one angle, and N = θ for the other,
we may write
θ + dθ + θ
θ + dθ − θ
· sin
,
2
2
dy = 2 cos(θ + 12 dθ) · sin 12 dθ.
dy = 2 cos
or,
But if we regard dθ as indefinitely small, then in the limit we may
neglect 21 dθ by comparison with θ, and may also take sin 12 dθ as being
the same as 12 dθ. The equation then becomes:
dy = 2 cos θ × 12 dθ;
dy = cos θ · dθ,
and, finally,
dy
= cos θ.
dθ
164
CALCULUS MADE EASY
The accompanying curves, Figs. 44 and 45, show, plotted to scale,
dy
= cos θ, for the corresponding values
the values of y = sin θ, and
dθ
of θ.
y
1
0.5
θ
30◦ 60◦ 90◦ 120◦ 150◦ 180◦
270◦
360◦
−0.5
−1
Fig. 44.
y
1
0.5
θ
30◦ 60◦ 90◦
180◦
−0.5
−1
Fig. 45.
Take next the cosine.
Let y = cos θ.
270◦
360◦
SINES AND COSINES
Now cos θ = sin
Therefore
π
2
165
−θ .
π
π
dy = d sin
− θ = cos
− θ × d(−θ),
2
2
π
− θ × (−dθ),
= cos
2
dy
π
= − cos
−θ .
dθ
2
And it follows that
dy
= − sin θ.
dθ
Lastly, take the tangent.
Let
y = tan θ,
dy = tan(θ + dθ) − tan θ.
Expanding, as shown in books on trigonometry,
tan θ + tan dθ
;
1 − tan θ · tan dθ
tan θ + tan dθ
− tan θ
dy =
1 − tan θ · tan dθ
(1 + tan2 θ) tan dθ
.
=
1 − tan θ · tan dθ
tan(θ + dθ) =
whence
Now remember that if dθ is indefinitely diminished, the value
of tan dθ becomes identical with dθ, and tan θ · dθ is negligibly small
CALCULUS MADE EASY
166
compared with 1, so that the expression reduces to
dy =
so that
or
(1 + tan2 θ) dθ
,
1
dy
= 1 + tan2 θ,
dθ
dy
= sec2 θ.
dθ
Collecting these results, we have:
y
dy
dθ
sin θ
cos θ
cos θ
− sin θ
tan θ
sec2 θ
Sometimes, in mechanical and physical questions, as, for example,
in simple harmonic motion and in wave-motions, we have to deal with
angles that increase in proportion to the time. Thus, if T be the time
of one complete period, or movement round the circle, then, since the
angle all round the circle is 2π radians, or 360◦ , the amount of angle
moved through in time t, will be
or
t
θ = 2π , in radians,
T
t
θ = 360 , in degrees.
T
If the frequency, or number of periods per second, be denoted by n,
1
then n = , and we may then write:
T
θ = 2πnt.
SINES AND COSINES
167
Then we shall have
y = sin 2πnt.
If, now, we wish to know how the sine varies with respect to time,
we must differentiate with respect, not to θ, but to t. For this we must
resort to the artifice explained in Chapter IX., p. 66, and put
dy dθ
dy
=
· .
dt
dθ dt
Now
dθ
will obviously be 2πn; so that
dt
dy
= cos θ × 2πn
dt
= 2πn · cos 2πnt.
Similarly, it follows that
d(cos 2πnt)
= −2πn · sin 2πnt.
dt
Second Differential Coefficient of Sine or Cosine.
We have seen that when sin θ is differentiated with respect to θ it
becomes cos θ; and that when cos θ is differentiated with respect to θ
it becomes − sin θ; or, in symbols,
d2 (sin θ)
= − sin θ.
dθ2
So we have this curious result that we have found a function such
that if we differentiate it twice over, we get the same thing from which
we started, but with the sign changed from + to −.
CALCULUS MADE EASY
168
The same thing is true for the cosine; for differentiating cos θ gives
us − sin θ, and differentiating − sin θ gives us − cos θ; or thus:
d2 (cos θ)
= − cos θ.
dθ2
Sines and cosines are the only functions of which the second differential coefficient is equal (and of opposite sign to) the original function.
Examples.
With what we have so far learned we can now differentiate expressions of a more complex nature.
(1) y = arc sin x.
If y is the arc whose sine is x, then x = sin y.
dx
= cos y.
dy
Passing now from the inverse function to the original one, we get
Now
hence
dy
1
1
=
=
.
dx
dx
cos y
dy
q
√
cos y = 1 − sin2 y = 1 − x2 ;
dy
1
=√
,
dx
1 − x2
a rather unexpected result.
(2) y = cos3 θ.
This is the same thing as y = (cos θ)3 .
SINES AND COSINES
Let cos θ = v; then y = v 3 ;
dy
= 3v 2 .
dv
dv
= − sin θ.
dθ
dy dv
dy
=
×
= −3 cos2 θ sin θ.
dθ
dv dθ
(3) y = sin(x + a).
Let x + a = v; then y = sin v.
dy
= cos v;
dv
dv
= 1 and
dx
dy
= cos(x + a).
dx
(4) y = log sin θ.
Let sin θ = v; y = log v.
dy
1
dv
= ;
= cos θ;
dv
v
dθ
dy
1
=
× cos θ = cot θ.
dθ
sin θ
(5) y = cot θ =
cos θ
.
sin θ
dy
− sin2 θ − cos2 θ
=
dθ
sin2 θ
= −(1 + cot2 θ) = − cosec2 θ.
(6) y = tan 3θ.
Let 3θ = v; y = tan v;
dy
= sec2 v.
dv
dv
= 3;
dθ
(7) y =
dy
= 3 sec2 3θ.
dθ
√
1
1 + 3 tan2 θ; y = (1 + 3 tan2 θ) 2 .
169
CALCULUS MADE EASY
Let 3 tan2 θ = v.
1
y = (1 + v) 2 ;
dy
1
= √
(see p. 67);
dv
2 1+v
dv
= 6 tan θ sec2 θ
dθ
(for, if tan θ = u,
v = 3u2 ;
hence
hence
dv
= 6u;
du
du
= sec2 θ;
dθ
dv
= 6(tan θ sec2 θ)
dθ
dy
6 tan θ sec2 θ
= √
.
dθ
2 1 + 3 tan2 θ
(8) y = sin x cos x.
dy
= sin x(− sin x) + cos x × cos x
dx
= cos2 x − sin2 x.
Exercises XIV.
(See page 261 for Answers.)
(1) Differentiate the following:
π
(i) y = A sin θ −
.
2
(ii) y = sin2 θ; and y = sin 2θ.
(iii) y = sin3 θ;
and y = sin 3θ.
(2) Find the value of θ for which sin θ × cos θ is a maximum.
1
(3) Differentiate y =
cos 2πnt.
2π
170
SINES AND COSINES
171
dy
.
dx
(5) Differentiate y = log cos x.
(4) If y = sin ax , find
(6) Differentiate y = 18.2 sin(x + 26◦ ).
(7) Plot the curve y = 100 sin(θ − 15◦ ); and show that the slope of
the curve at θ = 75◦ is half the maximum slope.
dy
(8) If y = sin θ · sin 2θ, find
.
dθ
(9) If y = a · tanm (θn ), find the differential coefficient of y with
respect to θ.
(10) Differentiate y = x sin2 x.
(11) Differentiate the three equations of Exercises XIII. (p. 160),
No. 4, and compare their differential coefficients, as to whether they
are equal, or nearly equal, for very small values of x, or for very large
values of x, or for values of x in the neighbourhood of x = 30.
(12) Differentiate the following:
(i) y = sec x.
(iii) y = arc tan x.
√
(v) y = tan x × 3 sec x.
(ii) y = arc cos x.
(iv) y = arc sec x.
(13) Differentiate y = sin(2θ + 3)2.3 .
(14) Differentiate y = θ3 + 3 sin(θ + 3) − 3sin θ − 3θ .
(15) Find the maximum or minimum of y = θ cos θ.
CHAPTER XVI.
PARTIAL DIFFERENTIATION.
We sometimes come across quantities that are functions of more than
one independent variable. Thus, we may find a case where y depends
on two other variable quantities, one of which we will call u and the
other v. In symbols
y = f (u, v).
Take the simplest concrete case.
Let
y = u × v.
What are we to do? If we were to treat v as a constant, and differentiate
with respect to u, we should get
dyv = v du;
or if we treat u as a constant, and differentiate with respect to v, we
should have:
dyu = u dv.
The little letters here put as subscripts are to show which quantity
has been taken as constant in the operation.
PARTIAL DIFFERENTIATION
173
Another way of indicating that the differentiation has been performed only partially, that is, has been performed only with respect to
one of the independent variables, is to write the differential coefficients
with Greek deltas, like ∂, instead of little d. In this way
∂y
= v,
∂u
∂y
= u.
∂v
If we put in these values for v and u respectively, we shall have

∂y


dyv =
du,
∂u
which are partial differentials.

∂y

dyu =
dv, 
∂v
But, if you think of it, you will observe that the total variation of y
depends on both these things at the same time. That is to say, if both
are varying, the real dy ought to be written
dy =
∂y
∂y
du +
dv;
∂u
∂v
and
this
is called
a total differential. In some books it is written dy =
dy
dy
du +
dv.
du
dv
Example (1). Find the partial differential coefficients of the expression w = 2ax2 + 3bxy + 4cy 3 . The answers are:

∂w

= 4ax + 3by. 

∂x
∂w

= 3bx + 12cy 2 .

∂y
CALCULUS MADE EASY
174
The first is obtained by supposing y constant, the second is obtained
by supposing x constant; then
dw = (4ax + 3by) dx + (3bx + 12cy 2 ) dy.
Example (2). Let z = xy . Then, treating first y and then x as
constant, we get in the usual way

∂z

y−1

= yx ,

∂x
∂z

= xy × log x,

∂y
so that dz = yxy−1 dx + xy log x dy.
Example (3). A cone having height h and radius of base r has
volume V = 31 πr2 h. If its height remains constant, while r changes,
the ratio of change of volume, with respect to radius, is different from
ratio of change of volume with respect to height which would occur if
the height were varied and the radius kept constant, for

2π 
∂V
=
rh,
∂r
3
∂V
π

= r2 . 
∂h
3
The variation when both the radius and the height change is given
2π
π
by dV =
rh dV + r2 dh.
3
3
Example (4). In the following example F and f denote two arbitrary functions of any form whatsoever. For example, they may be
sine-functions, or exponentials, or mere algebraic functions of the two
PARTIAL DIFFERENTIATION
175
independent variables, t and x. This being understood, let us take the
expression
y = F (x + at) + f (x − at),
or,
y = F (w) + f (v);
where
w = x + at,
Then
and v = x − at.
∂F (w) ∂w ∂f (v) ∂v
∂y
=
·
+
·
∂x
∂w
∂x
∂v
∂x
0
0
= F (w) · 1 + f (v) · 1
(where the figure 1 is simply the coefficient of x in w and v);
and
Also
and
whence
∂ 2y
= F 00 (w) + f 00 (v).
∂x2
∂y
∂F (w) ∂w ∂f (v) ∂v
=
·
+
·
∂t
∂w
∂t
∂v
∂t
0
0
= F (w) · a − f (v)a;
∂ 2y
= F 00 (w)a2 + f 00 (v)a2 ;
∂t2
2
∂ 2y
2 ∂ y
=
a
.
∂t2
∂x2
This differential equation is of immense importance in mathematical
physics.
Maxima and Minima of Functions of two Independent
Variables.
Example (5). Let us take up again Exercise IX., p. 107, No. 4.
Let x and y be the length of two of the portions of the string.
The third is 30 − (x + y), and the area of the triangle is A =
CALCULUS MADE EASY
176
p
s(s − x)(s − y)(s − 30 + x + y), where s is the half perimeter, 15, so
√
that A = 15P , where
P = (15 − x)(15 − y)(x + y − 15)
= xy 2 + x2 y − 15x2 − 15y 2 − 45xy + 450x + 450y − 3375.
Clearly A is maximum when P is maximum.
dP =
∂P
∂P
dx +
dy.
∂x
∂y
For a maximum (clearly it will not be a minimum in this case), one
must have simultaneously
∂P
= 0 and
∂x
∂P
= 0;
∂y

2xy − 30x + y 2 − 45y + 450 = 0,
that is,
2xy − 30y + x2 − 45x + 450 = 0.
An immediate solution is x = y.
If we now introduce this condition in the value of P , we find
P = (15 − x)2 (2x − 15) = 2x3 − 75x2 + 900x − 3375.
For maximum or minimum,
x = 15 or x = 10.
dP
= 6x2 − 150x + 900 = 0, which gives
dx
Clearly x = 15 gives minimum area; x = 10 gives the maximum, for
d2 P
= 12x − 150, which is +30 for x = 15 and −30 for x = 10.
dx2
PARTIAL DIFFERENTIATION
177
Example (6). Find the dimensions of an ordinary railway coal truck
with rectangular ends, so that, for a given volume V the area of sides
and floor together is as small as possible.
The truck is a rectangular box open at the top. Let x be the length
V
. The surface area is S =
and y be the width; then the depth is
xy
2V
2V
+
.
xy +
x
y
∂S
∂S
2V
2V
dS =
dx +
dy = y − 2 dx + x − 2 dy.
∂x
∂y
x
y
For minimum (clearly it won’t be a maximum here),
y−
2V
= 0,
x2
x−
2V
= 0.
y2
4V
,
Here also, an immediate solution is x = y, so that S = x2 +
x
dS
4V
= 2x − 2 = 0 for minimum, and
dx
x
√
3
x = 2V .
Exercises XV.
(See page 263 for Answers.)
x3
y
− 2x3 y − 2y 2 x + with respect
3
3
to x alone, and with respect to y alone.
(1) Differentiate the expression
(2) Find the partial differential coefficients with respect to x, y
and z, of the expression
x2 yz + xy 2 z + xyz 2 + x2 y 2 z 2 .
CALCULUS MADE EASY
178
(3) Let r2 = (x − a)2 + (y − b)2 + (z − c)2 .
Find the value of
∂ 2r ∂ 2r
∂r ∂r ∂r
+
+ . Also find the value of
+
+
∂x ∂y ∂z
∂x2 ∂y 2
∂ 2r
.
∂z 2
(4) Find the total differential of y = uv .
(5) Find the total differential of y = u3 sin v; of y = (sin x)u ; and of
log u
y=
.
v
(6) Verify that the sum of three quantities x, y, z, whose product is
a constant k, is maximum when these three quantities are equal.
(7) Find the maximum or minimum of the function
u = x + 2xy + y.
(8) The post-office regulations state that no parcel is to be of such
a size that its length plus its girth exceeds 6 feet. What is the greatest volume that can be sent by post (a) in the case of a package of
rectangular cross section; (b) in the case of a package of circular cross
section.
(9) Divide π into 3 parts such that the continued product of their
sines may be a maximum or minimum.
(10) Find the maximum or minimum of u =
x+y
.
xy
(11) Find maximum and minimum of
u = y + 2x − 2 log y − log x.
PARTIAL DIFFERENTIATION
179
(12) A telpherage bucket of given capacity has the shape of a horizontal isosceles triangular prism with the apex underneath, and the
opposite face open. Find its dimensions in order that the least amount
of iron sheet may be used in its construction.
CHAPTER XVII.
INTEGRATION.
TheZ great secret has already been revealed that this mysterious symbol , which is after all only a long S, merely means “the sum of,” or
“the sum of all such quantities as.” It therefore resembles that other
P
symbol
(the Greek Sigma), which is also a sign of summation. There
is this difference, however, in the practice of mathematical men as to
P
the use of these signs, that while
is generally used to
Z indicate the
sum of a number of finite quantities, the integral sign
is generally
used to indicate the summing up of a vast number of small quantities of
indefinitely minute magnitude,
Z mere elements
Z in fact, that go to make
up the total required. Thus dy = y, and dx = x.
Any one can understand how the whole of anything can be conceived
of as made up of a lot of little bits; and the smaller the bits the more of
them there will be. Thus, a line one inch long may be conceived as made
up of 10 pieces, each
1
100
1
10
of an inch long; or of 100 parts, each part being
of an inch long; or of 1, 000, 000 parts, each of which is
1
1,000,000
of
an inch long; or, pushing the thought to the limits of conceivability, it
may be regarded as made up of an infinite number of elements each of
which is infinitesimally small.
Yes, you will say, but what is the use of thinking of anything that
181
INTEGRATION
way? Why not think of it straight off, as a whole? The simple reason is
that there are a vast number of cases in which one cannot calculate the
bigness of the thing as a whole without reckoning up the sum of a lot
of small parts. The process of “integrating” is to enable us to calculate
totals that otherwise we should be unable to estimate directly.
Let us first take one or two simple cases to familiarize ourselves with
this notion of summing up a lot of separate parts.
Consider the series:
1 + 21 + 14 + 18 +
1
16
+
1
32
+
1
64
+ etc.
Here each member of the series is formed by taking it half the value
of the preceding. What is the value of the total if we could go on to
an infinite number of terms? Every schoolboy knows that the answer
is 2. Think of it, if you like, as a line. Begin with one inch; add a half
1/2
1
1/4
1/8
Fig. 46.
inch, add a quarter; add an eighth; and so on. If at any point of the
operation we stop, there will still be a piece wanting to make up the
whole 2 inches; and the piece wanting will always be the same size as
the last piece added. Thus, if after having put together 1, 12 , and 14 , we
stop, there will be
will still be
1
64
1
4
wanting. If we go on till we have added
1
,
64
there
wanting. The remainder needed will always be equal to
the last term added. By an infinite number of operations only should
we reach the actual 2 inches. Practically we should reach it when we
182
CALCULUS MADE EASY
got to pieces so small that they could not be drawn—that would be
after about 10 terms, for the eleventh term is
1
.
1024
If we want to go
so far that not even a Whitworth’s measuring machine would detect it,
we should merely have to go to about 20 terms. A microscope would
not show even the 18th term! So the infinite number of operations is no
such dreadful thing after all. The integral is simply the whole lot. But,
as we shall see, there are cases in which the integral calculus enables us
to get at the exact total that there would be as the result of an infinite
number of operations. In such cases the integral calculus gives us a
rapid and easy way of getting at a result that would otherwise require
an interminable lot of elaborate working out. So we had best lose no
time in learning how to integrate.
Slopes of Curves, and the Curves themselves.
Let us make a little preliminary enquiry about the slopes of curves.
For we have seen that differentiating a curve means finding an expression for its slope (or for its slopes at different points). Can we perform
the reverse process of reconstructing the whole curve if the slope (or
slopes) are prescribed for us?
Go back to case (2) on p. 82. Here we have the simplest of curves,
a sloping line with the equation
y = ax + b.
We know that here b represents the initial height of y when x = 0,
dy
, is the “slope” of the line. The line
and that a, which is the same as
dx
183
INTEGRATION
Y
b
O
X
Fig. 47.
has a constant slope. All along it the elementary triangles
dy
dx
have the same proportion between height and base. Suppose we were
to take the dx’s, and dy’s of finite magnitude, so that 10 dx’s made up
one inch, then there would be ten little triangles like
Now, suppose that we were ordered to reconstruct the “curve,”
dy
starting merely from the information that
= a. What could we
dx
do? Still taking the little d’s as of finite size, we could draw 10 of them,
all with the same slope, and then put them together, end to end, like
this: And, as the slope is the same for all, they would join to make, as
dy
in Fig. 48, a sloping line sloping with the correct slope
= a. And
dx
whether we take the dy’s and dx’s as finite or infinitely small, as they
y
are all alike, clearly = a, if we reckon y as the total of all the dy’s,
x
and x as the total of all the dx’s. But whereabouts are we to put this
184
CALCULUS MADE EASY
Y
c
O
X
Fig. 48.
sloping line? Are we to start at the origin O, or higher up? As the only
information we have is as to the slope, we are without any instructions
as to the particular height above O; in fact the initial height is undetermined. The slope will be the same, whatever the initial height. Let
us therefore make a shot at what may be wanted, and start the sloping
line at a height C above O. That is, we have the equation
y = ax + C.
It becomes evident now that in this case the added constant means
the particular value that y has when x = 0.
Now let us take a harder case, that of a line, the slope of which is
not constant, but turns up more and more. Let us assume that the
upward slope gets greater and greater in proportion as x grows. In
INTEGRATION
185
symbols this is:
dy
= ax.
dx
Or, to give a concrete case, take a = 15 , so that
dy
= 15 x.
dx
Then we had best begin by calculating a few of the values of the
slope at different values of x, and also draw little diagrams of them.
When
x = 0,
x = 1,
x = 2,
dy
= 0,
dx
dy
= 0.2,
dx
dy
= 0.4,
dx
x = 3,
dy
= 0.6,
dx
x = 4,
dy
= 0.8,
dx
x = 5,
dy
= 1.0.
dx
Now try to put the pieces together, setting each so that the middle
of its base is the proper distance to the right, and so that they fit
together at the corners; thus (Fig. 49). The result is, of course, not
a smooth curve: but it is an approximation to one. If we had taken
bits half as long, and twice as numerous, like Fig. 50, we should have a
186
CALCULUS MADE EASY
Y
O
P
1
2
3
4
5
X
Fig. 49.
better approximation. But for a perfect curve we ought to take each dx
and its corresponding dy infinitesimally small, and infinitely numerous.
P
Y
O
1
2
3
4
5
X
Fig. 50.
Then, how much ought the value of any y to be? Clearly, at any
point P of the curve, the value of y will be
Z the sum of all the little dy’s
from 0 up to that level, that is to say,
dy = y. And as each dy is
equal to 51 x · dx, it follows that the whole y will beZ equal to the sum of
1
all such bits as 15 x · dx, or, as we should write it,
x · dx.
5
Z
1
Now if x had been constant,
x · dx would have been the same
5
187
INTEGRATION
as
1
x
5
Z
dx, or
1 2
x.
5
But x began by being 0, and increases to the
particular value of x at the
Z point P , so that its average value from 0 to
that point is 21 x. Hence
1
x dx
5
=
1 2
x;
10
or y =
1 2
x.
10
But, as in the previous case, this requires the addition of an undetermined constant C, because we have not been told at what height
above the origin the curve will begin, when x = 0. So we write, as the
equation of the curve drawn in Fig. 51,
y=
1 2
x
10
+ C.
Y
y
C
x
Fig. 51.
Exercises XVI.
(See page 264 for Answers.)
(1) Find the ultimate sum of
2
3
+ 13 + 61 +
1
12
+
1
24
+ etc.
(2) Show that the series 1 − 21 + 13 − 41 + 15 − 61 + 17 etc., is convergent,
and find its sum to 8 terms.
CALCULUS MADE EASY
188
x2 x3 x4
+
−
+ etc., find log 1.3.
2
3
4
(4) Following a reasoning similar to that explained in this chapter,
(3) If log (1 + x) = x −
find y,
(a) if
(5) If
dy
dy
= 41 x; (b) if
= cos x.
dx
dx
dy
= 2x + 3, find y.
dx
CHAPTER XVIII.
INTEGRATING AS THE REVERSE OF
DIFFERENTIATING.
Differentiating is the process by which when y is given us (as a
dy
.
function of x), we can find
dx
Like every other mathematical operation, the process of differentiady
= 4x3 ;
tion may be reversed; thus, if differentiating y = x4 gives us
dx
dy
if one begins with
= 4x3 one would say that reversing the process
dx
would yield y = x4 . But here comes in a curious point. We should get
dy
= 4x3 if we had begun with any of the following: x4 , or x4 + a,
dx
or x4 + c, or x4 with any added constant. So it is clear that in working
dy
backwards from
to y, one must make provision for the possibility
dx
of there being an added constant, the value of which will be undetermined until ascertained in some other way. So, if differentiating xn
dy
yields nxn−1 , going backwards from
= nxn−1 will give us y = xn +C;
dx
where C stands for the yet undetermined possible constant.
Clearly, in dealing with powers of x, the rule for working backwards
will be: Increase the power by 1, then divide by that increased power,
and add the undetermined constant.
CALCULUS MADE EASY
So, in the case where
190
dy
= xn ,
dx
working backwards, we get
y=
1
xn+1 + C.
n+1
If differentiating the equation y = axn gives us
dy
= anxn−1 ,
dx
it is a matter of common sense that beginning with
dy
= anxn−1 ,
dx
and reversing the process, will give us
y = axn .
So, when we are dealing with a multiplying constant, we must simply
put the constant as a multiplier of the result of the integration.
dy
Thus, if
= 4x2 , the reverse process gives us y = 43 x3 .
dx
But this is incomplete. For we must remember that if we had started
with
y = axn + C,
where C is any constant quantity whatever, we should equally have
found
dy
= anxn−1 .
dx
So, therefore, when we reverse the process we must always remember
to add on this undetermined constant, even if we do not yet know what
its value will be.
HOW TO INTEGRATE
191
This process, the reverse of differentiating, is called integrating; for
it consists in finding the value of the whole quantity y when you are
dy
. Hitherto we have as much as
given only an expression for dy or for
dx
possible kept dy and dx together as a differential coefficient: henceforth
we shall more often have to separate them.
If we begin with a simple case,
dy
= x2 .
dx
We may write this, if we like, as
dy = x2 dx.
Now this is a “differential equation” which informs us that an element of y is equal to the corresponding element of x multiplied by x2 .
Now, what we want is the integral; therefore, write down with the
proper symbol the instructions to integrate both sides, thus:
Z
Z
dy = x2 dx.
[Note as to reading integrals: the above would be read thus:
“Integral dee-wy equals integral eks-squared dee-eks.”]
We haven’t yet integrated: we have only written down instructions
to integrate—if we can. Let us try. Plenty of other fools can do it—why
not we also? The left-hand side is simplicity itself. The sum of all the
bits of y is the same thing as y itself. So we may at once put:
Z
y = x2 dx.
CALCULUS MADE EASY
192
But when we come to the right-hand side of the equation we must
remember that what we have got to sum up together is not all theZdx’s,
but all such terms as x2 dx; and this will not be the same as x2
dx,
because x2 is not a constant. For some of the dx’s will be multiplied
by big values of x2 , and some will be multiplied by small values of x2 ,
according to what x happens to be. So we must bethink ourselves as
to what we know about this process of integration being the reverse
of differentiation. Now, our rule for this reversed process—see p. 189
ante—when dealing with xn is “increase the power by one, and divide
by the same number as this increased power.” That is to say, x2 dx will
be changed∗ to 31 x3 . Put this into the equation; but don’t forget to add
the “constant of integration” C at the end. So we get:
y = 31 x3 + C.
You have actually performed the integration. How easy!
Let us try another simple case.
Let
dy
= ax12 ,
dx
where a is any constant multiplier. Well, we found when differentiating (see p. 27) that any constant factor in the value of y reappeared
∗
You may ask, what has become of the little dx at the end? Well, remember
that it was really part of the differential coefficient, and when changed over to the
right-hand side, as in the x2 dx, serves as a reminder that x is the independent
variable with respect to which the operation is to be effected; and, as the result of
the product being totalled up, the power of x has increased by one. You will soon
become familiar with all this.
HOW TO INTEGRATE
193
dy
. In the reversed process of integrating, it
dx
will therefore also reappear in the value of y. So we may go to work as
unchanged in the value of
before, thus
dy = ax12 · dx,
Z
Z
dy = ax12 · dx,
Z
Z
dy = a x12 dx,
y =a×
1 13
x
13
+ C.
So that is done. How easy!
We begin to realize now that integrating is a process of finding our
way back, as compared with differentiating. If ever, during differentiating, we have found any particular expression—in this example ax12 —we
can find our way back to the y from which it was derived. The contrast
between the two processes may be illustrated by the following remark
due to a well-known teacher. If a stranger were set down in Trafalgar
Square, and told to find his way to Euston Station, he might find the
task hopeless. But if he had previously been personally conducted from
Euston Station to Trafalgar Square, it would be comparatively easy to
him to find his way back to Euston Station.
Integration of the Sum or Difference of two Functions.
Let
then
dy
= x2 + x3 ,
dx
dy = x2 dx + x3 dx.
CALCULUS MADE EASY
194
There is no reason why we should not integrate each term separately:
for, as may be seen on p. 34, we found that when we differentiated the
sum of two separate functions, the differential coefficient was simply the
sum of the two separate differentiations. So, when we work backwards,
integrating, the integration will be simply the sum of the two separate
integrations.
Our instructions will then be:
Z
Z
dy = (x2 + x3 ) dx
Z
Z
2
= x dx + x3 dx
y = 13 x3 + 14 x4 + C.
If either of the terms had been a negative quantity, the corresponding term in the integral would have also been negative. So that differences are as readily dealt with as sums.
How to deal with Constant Terms.
Suppose there is in the expression to be integrated a constant term—
such as this:
dy
= xn + b.
dx
This is laughably easy. For you have only to remember that when
dy
you differentiated the expression y = ax, the result was
= a. Hence,
dx
when you work the other way and integrate, the constant reappears
195
EASIEST INTEGRATIONS
multiplied by x. So we get
Z
dy = xn dx + b · dx,
Z
Z
n
dy = x dx + b dx,
y=
1
xn+1 + bx + C.
n+1
Here are a lot of examples on which to try your newly acquired
powers.
Examples.
dy
(1) Given
= 24x11 . Find y.
Ans. y = 2x12 + C.
Z dx
Z
(2) Find (a + b)(x + 1) dx.
It is (a + b) (x + 1) dx
2
Z
Z
x
x dx + dx
or (a + b)
or (a + b)
+ x + C.
2
du
1
3
(3) Given
= gt 2 . Find u.
Ans. u = 23 gt 2 + C.
dt
dy
(4)
= x3 − x2 + x. Find y.
dx
dy = (x3 − x2 + x) dx or
3
2
dy = x dx − x dx + x dx;
and
Z
y=
3
x dx −
Z
2
x dx +
y = 14 x4 − 31 x3 + 12 x2 + C.
(5) Integrate 9.75x2.25 dx.
Ans. y = 3x3.25 + C.
Z
x dx;
CALCULUS MADE EASY
196
All these are easy enough. Let us try another case.
dy
= ax−1 .
dx
Let
Proceeding as before, we will write
Z
Z
−1
dy = ax · dx,
dy = a x−1 dx.
Well, but what is the integral of x−1 dx?
If you look back amongst the results of differentiating x2 and x3
and xn , etc., you will find we never got x−1 from any one of them as
dy
. We got 3x2 from x3 ; we got 2x from x2 ; we got 1
the value of
dx
from x1 (that is, from x itself); but we did not get x−1 from x0 , for
two very good reasons. First, x0 is simply = 1, and is a constant, and
could not have a differential coefficient. Secondly, even if it could be
differentiated, its differential coefficient (got by slavishly following the
usual rule) would be 0 × x−1 , and that multiplication by zero gives it
zero value! Therefore when we now come to try to integrate x−1 dx, we
see that it does not come in anywhere in the powers of x that are given
by the rule:
Z
xn dx =
1
xn+1 .
n+1
It is an exceptional case.
Well; but try again. Look through all the various differentials obtained from various functions of x, and try to find amongst them x−1 .
dy
A sufficient search will show that we actually did get
= x−1 as the
dx
result of differentiating the function y = log x (see p. 145).
EASIEST INTEGRATIONS
197
Then, of course, since we know that differentiating log x gives
us x−1 , we know that, by reversing the process, integrating dy = x−1 dx
will give us y = log x. But we must not forget the constant factor a
that was given, nor must we omit to add the undetermined constant of
integration. This then gives us as the solution to the present problem,
y = a log x + C.
N.B.—Here note this very remarkable fact, that we could not have
integrated in the above case if we had not happened to know the corresponding differentiation. If no one had found out that differentiating
log x gave x−1 , we should have been utterly stuck by the problem how
to integrate x−1 dx. Indeed it should be frankly admitted that this is
one of the curious features of the integral calculus:—that you can’t integrate anything before the reverse process of differentiating something
else has yielded that expression which you want to integrate. No one,
even to-day, is able to find the general integral of the expression,
dy
2
= a−x ,
dx
2
because a−x has never yet been found to result from differentiating
anything else.
Another
Z simple case.
Find (x + 1)(x + 2) dx.
On looking at the function to be integrated, you remark that it
is the product of two different functions of x. You could, you think,
integrate (x+1) dx by itself, or (x+2) dx by itself. Of course you could.
But what to do with a product? None of the differentiations you have
CALCULUS MADE EASY
198
learned have yielded you for the differential coefficient a product like
this. Failing such, the simplest thing is to multiply up the two functions,
and then integrate. This gives us
Z
(x2 + 3x + 2) dx.
And this is the same as
Z
Z
Z
2
x dx + 3x dx + 2 dx.
And performing the integrations, we get
1 3
x
3
+ 32 x2 + 2x + C.
Some other Integrals.
Now that we know that integration is the reverse of differentiation,
we may at once look up the differential coefficients we already know, and
see from what functions they were derived. This gives us the following
integrals ready made:
x
−1
Z
(p. 145);
1
(p. 145);
x+a
x
−x
Z
Z
(p. 139);
Z
x−1 dx
= log x + C.
1
dx = log (x + a) + C.
x+a
x dx
= x + C.
−x dx
= −−x + C
SOME OTHER INTEGRALS
199
x × 0 − 1 × x
dy
=−
= −x ).
dx
2x
Z
sin x (p. 164);
sin x dx = − cos x + C.
Z
cos x (p. 163);
cos x dx = sin x + C.
1
,
x
(for if y = −
Also we may deduce the following:
Z
log x;
log x dx = x(log x − 1) + C
(for if y = x log x − x,
log10 x;
ax
cos ax;
(p. 146);
x
dy
= + log x − 1 = log x).
dx
x
Z
log10 x dx = 0.4343x(log x − 1) + C.
Z
ax
ax dx
=
+ C.
log a
Z
1
cos ax dx = sin ax + C
a
dy
(for if y = sin ax,
= a cos ax; hence to get cos ax one must differendx
1
tiate y = sin ax).
a
Z
1
sin ax;
sin ax dx = − cos ax + C.
a
Try also cos2 θ; a little dodge will simplify matters:
cos 2θ = cos2 θ − sin2 θ = 2 cos2 θ − 1;
hence
cos2 θ = 21 (cos 2θ + 1),
CALCULUS MADE EASY
Z
and
2
cos θ dθ =
1
2
200
Z
(cos 2θ + 1) dθ
Z
Z
1
1
= 2 cos 2θ dθ + 2 dθ.
=
sin 2θ θ
+ + C. (See also p. 225).
4
2
See also the Table of Standard Forms on pp. 249–251. You should
make such a table for yourself, putting in it only the general functions
which you have successfully differentiated and integrated. See to it that
it grows steadily!
On Double and Triple Integrals.
In many cases it is necessary to integrate some expression for two or
more variables contained in it; and in that case the sign of integration
appears more than once. Thus,
ZZ
f (x, y, ) dx dy
means that some function of the variables x and y has to be integrated
for each. It does not matter in which order they are done. Thus, take
the function x2 + y 2 . Integrating it with respect to x gives us:
Z
(x2 + y 2 ) dx = 13 x3 + xy 2 .
Now, integrate this with respect to y:
Z
( 31 x3 + xy 2 ) dy = 13 x3 y + 13 xy 3 ,
DOUBLE INTEGRALS
201
to which of course a constant is to be added. If we had reversed the
order of the operations, the result would have been the same.
In dealing with areas of surfaces and of solids, we have often to
integrate both for length and breadth, and thus have integrals of the
form
ZZ
u · dx dy,
where u is some property that depends, at each point, on x and on y.
This would then be called a surface-integral. It indicates that the value
of all such elements as u · dx · dy (that is to say, of the value of u over
a little rectangle dx long and dy broad) has to be summed up over the
whole length and whole breadth.
Similarly in the case of solids, where we deal with three dimensions. Consider any element of volume, the small cube whose dimensions are dx dy dz. If the figure of the solid be expressed by the function
f (x, y, z), then the whole solid will have the volume-integral,
ZZZ
volume =
f (x, y, z) · dx · dy · dz.
Naturally, such integrations have to be taken between appropriate limits∗ in each dimension; and the integration cannot be performed unless
one knows in what way the boundaries of the surface depend on x, y,
and z. If the limits for x are from x1 to x2 , those for y from y1 to y2 ,
and those for z from z1 to z2 , then clearly we have
Z z2 Z y2 Z x2
volume =
f (x, y, z) · dx · dy · dz.
z1
∗
y1
x1
See p. 206 for integration between limits.
202
CALCULUS MADE EASY
There are of course plenty of complicated and difficult cases; but, in
general, it is quite easy to see the significance of the symbols where they
are intended to indicate that a certain integration has to be performed
over a given surface, or throughout a given solid space.
Exercises XVII. (See p. 264 for the Answers.)
Z
(1) Find y dx when y 2 = 4ax.
Z
(2) Find
Z
(4) Find
Z
(6) Find
1 3
x dx.
a
3
dx.
x4
(3) Find
(x2 + a) dx.
(5) Integrate 5x− 2 .
Z
7
(4x3 + 3x2 + 2x + 1) dx.
dy
ax bx2 cx3
=
+
+
; find y.
dx
2
3
4
Z
Z 2
x +a
(9) Find (x + 3)3 dx.
dx.
(8) Find
x+a
Z
(10) Find (x + 2)(x − a) dx.
Z
√
√
(11) Find ( x + 3 x)3a2 dx.
Z
dθ
(12) Find (sin θ − 21 ) .
3
Z
Z
2
(13) Find cos aθ dθ.
(14) Find sin2 θ dθ.
(7) If
203
SIMPLE INTEGRATIONS
Z
(15) Find
sin aθ dθ.
Z
(17) Find
2
dx
.
1+x
Z
(16) Find
Z
(18) Find
3x dx.
dx
.
1−x
CHAPTER XIX.
ON FINDING AREAS BY INTEGRATING.
One use of the integral calculus is to enable us to ascertain the values
of areas bounded by curves.
Let us try to get at the subject bit by bit.
Y
Q
B
P
A
y2
y1
M
O
N
x1
x2
Fig. 52.
Let AB (Fig. 52) be a curve, the equation to which is known. That
is, y in this curve is some known function of x. Think of a piece of the
curve from the point P to the point Q.
FINDING AREAS BY INTEGRATING
205
Let a perpendicular P M be dropped from P , and another QN from
the point Q. Then call OM = x1 and ON = x2 , and the ordinates
P M = y1 and QN = y2 . We have thus marked out the area P QN M
that lies beneath the piece P Q. The problem is, how can we calculate
the value of this area?
The secret of solving this problem is to conceive the area as being divided up into a lot of narrow strips, each of them being of the
width dx. The smaller we take dx, the more of them there will be
between x1 and x2 . Now, the whole area is clearly equal to the sum of
the areas of all such strips. Our business will then be to discover an
expression for the area of any one narrow strip, and to integrate it so
as to add together all the strips. Now think of any one of the strips. It
will be like this: being bounded between two vertical sides,
with a flat bottom dx, and with a slightly curved sloping top.
Suppose we take its average height as being y; then, as its
width is dx, its area will be y dx. And seeing that we may take
the width as narrow as we please, if we only take it narrow
enough its average height will be the same as the height at the
middle of it. Now let us call the unknown value of the whole
area S, meaning surface. The area of one strip will be simply a bit of
the whole area, and may therefore be called dS. So we may write
area of 1 strip = dS = y · dx.
If then we add up all the strips, we get
Z
Z
total area S = dS = y dx.
CALCULUS MADE EASY
206
So then our finding S depends on whether we can integrate y ·dx for
the particular case, when we know what the value of y is as a function
of x.
For instance, if you were told that for the particular curve in question y = b + ax2 , no doubt Zyou could put that value into the expression
and say: then I must find
(b + ax2 ) dx.
That is all very well; but a little thought will show you that something more must be done. Because the area we are trying to find is not
the area under the whole length of the curve, but only the area limited
on the left by P M , and on the right by QN , it follows that we must
do something to define our area between those ‘limits.’
This introduces us to a new notion, namely that of integrating between limits. We suppose x to vary, and for the present purpose we do
not require any value of x below x1 (that is OM ), nor any value of x
above x2 (that is ON ). When an integral is to be thus defined between
two limits, we call the lower of the two values the inferior limit, and the
upper value the superior limit. Any integral so limited we designate as
a definite integral, by way of distinguishing it from a general integral
to which no limits are assigned.
In the symbols which give instructions to integrate, the limits are
marked by putting them at the top and bottom respectively of the sign
of integration. Thus the instruction
Z x=x2
x=x1
y · dx
will be read: find the integral of y · dx between the inferior limit x1 and
the superior limit x2 .
FINDING AREAS BY INTEGRATING
207
Sometimes the thing is written more simply
Z x2
y · dx.
x1
Well, but how do you find an integral between limits, when you have
got these instructions?
Look again at Fig. 52 (p. 204). Suppose we could find the area under
the larger piece of curve from A to Q, that is from x = 0 to x = x2 ,
naming the area AQN O. Then, suppose we could find the area under
the smaller piece from A to P , that is from x = 0 to x = x1 , namely
the area AP M O. If then we were to subtract the smaller area from
the larger, we should have left as a remainder the area P QN M , which
is what we want. Here we have the clue as to what to do; the definite
integral between the two limits is the difference between the integral
worked out for the superior limit and the integral worked out for the
lower limit.
Let us then go ahead. First, find the general integral thus:
Z
y dx,
and, as y = b + ax2 is the equation to the curve (Fig. 52),
Z
(b + ax2 ) dx
is the general integral which we must find.
Doing the integration in question by the rule (p. 193), we get
a
bx + x3 + C;
3
and this will be the whole area from 0 up to any value of x that we
may assign.
CALCULUS MADE EASY
208
Therefore, the larger area up to the superior limit x2 will be
a
bx2 + x32 + C;
3
and the smaller area up to the inferior limit x1 will be
a
bx1 + x31 + C.
3
Now, subtract the smaller from the larger, and we get for the area S
the value,
a
area S = b(x2 − x1 ) + (x32 − x31 ).
3
This is the answer we wanted. Let us give some numerical values.
Suppose b = 10, a = 0.06, and x2 = 8 and x1 = 6. Then the area S is
equal to
0.06 3
(8 − 63 )
3
= 20 + 0.02(512 − 216)
10(8 − 6) +
= 20 + 0.02 × 296
= 20 + 5.92
= 25.92.
Let us here put down a symbolic way of stating what we have ascertained about limits:
Z
x=x2
x=x1
y dx = y2 − y1 ,
where y2 is the integrated value of y dx corresponding to x2 , and y1 that
corresponding to x1 .
FINDING AREAS BY INTEGRATING
209
All integration between limits requires the difference between two
values to be thus found. Also note that, in making the subtraction the
added constant C has disappeared.
Examples.
(1) To familiarize ourselves with the process, let us take a case of
which we know the answer beforehand. Let us find the area of the
triangle (Fig. 53), which has base x = 12 and height y = 4. We know
beforehand, from obvious mensuration, that the answer will come 24.
y
4
x
O
12
Fig. 53.
Now, here we have as the “curve” a sloping line for which the equation is
y=
x
.
3
The area in question will be
Z x=12
Z
y · dx =
x=0
x=12
x=0
x
· dx.
3
x
dx (p. 192), and putting down the value of the general
3
integral in square brackets with the limits marked above and below, we
Integrating
210
CALCULUS MADE EASY
get
x=12
1 1 2
· x
+C
area =
3 2
x=0
2 x=12
x
+C
=
6 x=0
2 2
12
0
=
−
6
6
144
= 24. Ans.
=
6
Let us satisfy ourselves about this rather surprising dodge of calculation, by testing it on a simple example. Get some squared paper,
preferably some that is ruled in little squares of one-eighth inch or oney
5
4
3
2
1
3
O
6
9
12
x
Fig. 54.
tenth inch each way. On this squared paper plot out the graph of the
equation,
y=
x
.
3
The values to be plotted will be:
x
0
3
6
9
12
y
0
1
2
3
4
FINDING AREAS BY INTEGRATING
211
The plot is given in Fig. 54.
Now reckon out the area beneath the curve by counting the little
squares below the line, from x = 0 as far as x = 12 on the right. There
are 18 whole squares and four triangles, each of which has an area equal
to 1 12 squares; or, in total, 24 squares. Hence 24 is the numerical value
x
of the integral of dx between the lower limit of x = 0 and the higher
3
limit of x = 12.
As a further exercise, show that the value of the same integral between the limits of x = 3 and x = 15 is 36.
Y
b
a
a
O
x
X
Fig. 55.
(2) Find the area, between limits x = x1 and x = 0, of the curve
b
.
y=
x+a
Z x=x1
Z x=x1
b
Area =
y · dx =
dx
x+a
x=0
x=0
x
= b log (x + a) 0 1 + C
= b log (x1 + a) − log (0 + a)
CALCULUS MADE EASY
212
x1 + a
. Ans.
a
N.B.—Notice that in dealing with definite integrals the constant C
= b log
always disappears by subtraction.
Let it be noted that this process of subtracting one part from a
larger to find the difference is really a common practice. How do you
find the area of a plane ring (Fig. 56), the outer radius of which is r2
r2
r1
Fig. 56.
and the inner radius is r1 ? You know from mensuration that the area
of the outer circle is πr22 ; then you find the area of the inner circle, πr12 ;
then you subtract the latter from the former, and find area of ring
= π(r22 − r12 ); which may be written
π(r2 + r1 )(r2 − r1 )
= mean circumference of ring × width of ring.
(3) Here’s another case—that of the die-away curve (p. 153). Find
the area between x = 0 and x = a, of the curve (Fig. 57) whose equation
is
y = b−x .
Z x=a
Area = b
−x · dx.
x=0
213
FINDING AREAS BY INTEGRATING
The integration (p. 198) gives
a
= b −−x 0
= b −−a − (−−0 )
= b(1 − −a ).
p
Y
p1
b
p2
O
O
v1
X
a
v
v2
Fig. 58.
Fig. 57.
(4) Another example is afforded by the adiabatic curve of a perfect
gas, the equation to which is pv n = c, where p stands for pressure, v for
volume, and n is of the value 1.42 (Fig. 58).
Find the area under the curve (which is proportional to the work
done in suddenly compressing the gas) from volume v2 to volume v1 .
Here we have
Z
v=v2
cv −n · dv
v=v1
v2
1
1−n
=c
v
1−n
v1
1
=c
(v 1−n − v11−n )
1 −
n 2
−c
1
1
=
−
.
0.42 v20.42 v10.42
area =
214
CALCULUS MADE EASY
An Exercise.
Prove the ordinary mensuration formula, that the area A of a circle
whose radius is R, is equal to πR2 .
dr
r
R
Fig. 59.
Consider an elementary zone or annulus of the surface (Fig. 59), of
breadth dr, situated at a distance r from the centre. We may consider
the entire surface as consisting of such narrow zones, and the whole
area A will simply be the integral of all such elementary zones from
centre to margin, that is, integrated from r = 0 to r = R.
We have therefore to find an expression for the elementary area dA
of the narrow zone. Think of it as a strip of breadth dr, and of a length
that is the periphery of the circle of radius r, that is, a length of 2πr.
Then we have, as the area of the narrow zone,
dA = 2πr dr.
Hence the area of the whole circle will be:
Z
Z r=R
Z
A = dA =
2πr · dr = 2π
r=0
r=R
r=0
r · dr.
FINDING AREAS BY INTEGRATING
215
Now, the general integral of r · dr is 21 r2 . Therefore,
1 2 r=R
r r=0 ;
2
1 2 1 2
A = 2π 2 R − 2 (0) ;
A = 2π
or
A = πR2 .
whence
Another Exercise.
Let us find the mean ordinate of the positive part of the curve
y = x − x2 , which is shown in Fig. 60. To find the mean ordinate, we
Y
M
1/4
N
O
1
Fig. 60.
shall have to find the area of the piece OM N , and then divide it by
the length of the base ON . But before we can find the area we must
ascertain the length of the base, so as to know up to what limit we are
to integrate. At N the ordinate y has zero value; therefore, we must
look at the equation and see what value of x will make y = 0. Now,
clearly, if x is 0, y will also be 0, the curve passing through the origin O;
but also, if x = 1, y = 0; so that x = 1 gives us the position of the
point N .
CALCULUS MADE EASY
216
Then the area wanted is
Z
x=1
=
x=0
(x − x2 ) dx
2
1 3 1
x
−
x
2
3
0
1 1
= 2 − 3 − [0 − 0]
=
1
= 61 .
But the base length is 1.
Therefore, the average ordinate of the curve = 61 .
[N.B.—It will be a pretty and simple exercise in maxima and minima
to find by differentiation what is the height of the maximum ordinate.
It must be greater than the average.]
The mean ordinate of any curve, over a range from x = 0 to x = x1 ,
is given by the expression,
Z
1 x=x1
y · dx.
mean y =
x1 x=0
One can also find in the same way the surface area of a solid of
revolution.
Example.
The curve y = x2 − 5 is revolving about the axis of x. Find the area
of the surface generated by the curve between x = 0 and x = 6.
A point on the curve, the ordinate of which is y, describes a circumference of length 2πy, and a narrow belt of the surface, of width dx,
corresponding to this point, has for area 2πy dx. The total area is
3
6
Z x=6
Z x=6
x
2
2π
y dx = 2π
(x − 5) dx = 2π
− 5x
3
x=0
x=0
0
= 6.28 × 42 = 263.76.
FINDING AREAS BY INTEGRATING
217
Areas in Polar Coordinates.
When the equation of the boundary of an area is given as a function
of the distance r of a point of it from a fixed point O (see Fig. 61) called
the pole, and of the angle which r makes with the positive horizontal
A
dθ
B
r
θ
O
X
Fig. 61.
direction OX, the process just explained can be applied just as easily,
with a small modification. Instead of a strip of area, we consider a
small triangle OAB, the angle at O being dθ, and we find the sum of
all the little triangles making up the required area.
AB
The area of such a small triangle is approximately
× r or
2
r dθ
× r; hence the portion of the area included between the curve and
2
two positions of r corresponding to the angles θ1 and θ2 is given by
Z θ=θ2
1
r2 dθ.
2
θ=θ1
Examples.
(1) Find the area of the sector of 1 radian in a circumference of
radius a inches.
218
CALCULUS MADE EASY
The polar equation of the circumference is evidently r = a. The
area is
1
2
Z
θ=θ2
θ=θ1
a2
a dθ =
2
2
Z
θ=1
dθ =
θ=0
a2
.
2
(2) Find the area of the first quadrant of the curve (known as “Pascal’s Snail”), the polar equation of which is r = a(1 + cos θ).
Z θ= π
2
1
Area = 2
a2 (1 + cos θ)2 dθ
θ=0
2 Z θ= π
2
a
(1 + 2 cos θ + cos2 θ) dθ
=
2 θ=0
π
a2
θ sin 2θ 2
=
θ + 2 sin θ + +
2
2
4
0
2
a (3π + 8)
.
=
8
Volumes by Integration.
What we have done with the area of a little strip of a surface, we
can, of course, just as easily do with the volume of a little strip of a
solid. We can add up all the little strips that make up the total solid,
and find its volume, just as we have added up all the small little bits
that made up an area to find the final area of the figure operated upon.
Examples.
(1) Find the volume of a sphere of radius r.
A thin spherical shell has for volume 4πx2 dx (see Fig. 59, p. 214);
summing up all the concentric shells which make up the sphere, we
FINDING AREAS BY INTEGRATING
have
Z
x=r
x3
4πx dx = 4π
3
2
volume sphere =
x=0
r
219
= 43 πr3 .
0
y
dx
y
x
O
x
Fig. 62.
We can also proceed as follows: a slice of the sphere, of thickness dx,
has for volume πy 2 dx (see Fig. 62). Also x and y are related by the
expression
y 2 = r 2 − x2 .
Z
x=r
π(r2 − x2 ) dx
x=0
Z x=r
Z x=r
2
2
= 2π
r dx −
x dx
x=0
x=0
r
x3
4π 3
2
= 2π r x −
=
r .
3 0
3
Hence volume sphere = 2
(2) Find the volume of the solid generated by the revolution of the
curve y 2 = 6x about the axis of x, between x = 0 and x = 4.
CALCULUS MADE EASY
220
The volume of a strip of the solid is πy 2 dx.
Z x=4
Z x=4
2
x dx
πy dx = 6π
Hence
volume =
x=0
x=0
2 4
x
= 6π
= 48π = 150.8.
2 0
On Quadratic Means.
In certain branches of physics, particularly in the study of alternating electric currents, it is necessary to be able to calculate the quadratic
mean of a variable quantity. By “quadratic mean” is denoted the square
root of the mean of the squares of all the values between the limits considered. Other names for the quadratic mean of any quantity are its
“virtual” value, or its “r.m.s.” (meaning root-mean-square) value. The
French term is valeur efficace. If y is the function under consideration,
and the quadratic mean is to be taken between the limits of x = 0 and
x = l; then the quadratic mean is expressed as
s
Z l
2 1
y 2 dx.
l 0
Examples.
(1) To find the quadratic mean of the function y = ax (Fig. 63).
Z l
Here the integral is
a2 x2 dx, which is 31 a2 l3 .
0
Dividing by l and taking the square root, we have
1
quadratic mean = √ al.
3
221
FINDING AREAS BY INTEGRATING
Y
y
O
X
l
Fig. 63.
Here the arithmetical mean is 12 al; and the ratio of quadratic to
2
arithmetical mean (this ratio is called the form-factor ) is √ = 1.155.
3
(2) To find the quadratic mean of the function y = xa .
Z x=l
l2a+1
The integral is
x2a dx, that is
.
2a + 1
x=0
s
l2a
2
.
Hence
quadratic mean =
2a + 1
x
(3) To find the quadratic mean of the function y = a 2 .
Z x=l
Z x=l
x 2
The integral is
(a 2 ) dx, that is
ax dx,
x=0
x=0
or
which is
ax
log a
al − 1
.
log a
s
Hence the quadratic mean is
Exercises XVIII.
2
x=l
,
x=0
al − 1
.
l log a
(See p. 265 for Answers.)
(1) Find the area of the curve y = x2 + x − 5 between x = 0 and
x = 6, and the mean ordinates between these limits.
222
CALCULUS MADE EASY
√
(2) Find the area of the parabola y = 2a x between x = 0 and
x = a. Show that it is two-thirds of the rectangle of the limiting
ordinate and of its abscissa.
(3) Find the area of the positive portion of a sine curve and the
mean ordinate.
(4) Find the area of the positive portion of the curve y = sin2 x, and
find the mean ordinate.
(5) Find the area included between the two branches of the curve
5
y = x2 ± x 2 from x = 0 to x = 1, also the area of the positive portion
of the lower branch of the curve (see Fig. 30, p. 106).
(6) Find the volume of a cone of radius of base r, and of height h.
(7) Find the area of the curve y = x3 − log x between x = 0 and
x = 1.
(8) Find the volume generated by the curve y =
√
1 + x2 , as it
revolves about the axis of x, between x = 0 and x = 4.
(9) Find the volume generated by a sine curve revolving about the
axis of x. Find also the area of its surface.
(10) Find the area of the portion of the curve xy = a included between x = 1 and x = a. Find the mean ordinate between these limits.
(11) Show that the quadratic mean
√ of the function y = sin x, between
2
the limits of 0 and π radians, is
. Find also the arithmetical mean
2
of the same function between the same limits; and show that the formfactor is = 1.11.
FINDING AREAS BY INTEGRATING
223
(12) Find the arithmetical and quadratic means of the function x2 +
3x + 2, from x = 0 to x = 3.
(13) Find the quadratic mean and the arithmetical mean of the function y = A1 sin x + A1 sin 3x.
(14) A certain curve has the equation y = 3.420.21x . Find the area
included between the curve and the axis of x, from the ordinate at x = 2
to the ordinate at x = 8. Find also the height of the mean ordinate of
the curve between these points.
(15) Show that the radius of a circle, the area of which is twice the
area of a polar diagram, is equal to the quadratic mean of all the values
of r for that polar diagram.
(16) Find the volume generated by the curve y = ±
rotating about the axis of x.
xp
x(10 − x)
6
CHAPTER XX.
DODGES, PITFALLS, AND TRIUMPHS.
Dodges. A great part of the labour of integrating things consists in
licking them into some shape that can be integrated. The books—and
by this is meant the serious books—on the Integral Calculus are full of
plans and methods and dodges and artifices for this kind of work. The
following are a few of them.
Integration by Parts. This name is given to a dodge, the formula for
which is
Z
u dx = ux −
Z
x du + C.
It is useful in some
Z cases that you can’t tackle
Z directly, for it shows
that if in any case x du can be found, then u dx can also be found.
The formula can be deduced as follows. From p. 37, we have,
d(ux) = u dx + x du,
which may be written
u(dx) = d(ux) − x du,
which by direct integration gives the above expression.
DODGES, PITFALLS, AND TRIUMPHS
Examples.Z
(1) Find w · sin w dw.
225
Write u = w,
Z and for sin w · dw write dx. We shall then have
du = dw, while sin w · dw = − cos w = x.
Putting these into the formula, we get
Z
Z
w · sin w dw = w(− cos w) − − cos w dw
= −w cos w + sin w + C.
Z
xx dx.
(2) Find
Write
x dx = dv;
u = x,
then
v = x ,
du = dx,
Z
and
x
x
x dx = x −
Z
x dx (by the formula)
= xx − x = x (x − 1) + C.
Z
(3) Try
cos2 θ dθ.
u = cos θ,
cos θ dθ = dv.
du = − sin θ dθ,
Hence
Z
2
v = sin θ,
Z
sin2 θ dθ
Z
2 cos θ sin θ
+ (1 − cos2 θ) dθ
=
2
Z
Z
sin 2θ
=
+ dθ − cos2 θ dθ.
2
cos θ dθ = cos θ sin θ +
CALCULUS MADE EASY
sin 2θ
+θ
2
Z
sin 2θ θ
+ + C.
cos2 θ dθ =
4
2
Z
Hence
2
and
Z
(4) Find
226
cos2 θ dθ =
x2 sin x dx.
x2 = u,
Write
then
sin x dx = dv;
v = − cos x,
du = 2x dx,
Z
2
2
x sin x dx = −x cos x + 2
Z
x cos x dx.
Z
Now find
x cos x dx, integrating by parts (as in Example 1 above):
Z
x cos x dx = x sin x + cos x + C.
Hence
Z
x2 sin x dx = −x2 cos x + 2x sin x + 2 cos x + C 0
x2
= 2 x sin x + cos x 1 −
+ C 0.
2
Z √
(5) Find
1 − x2 dx.
Write
then
u=
√
1 − x2 ,
du = − √
x dx
1 − x2
dx = dv;
(see Chap. IX., p. 66)
DODGES, PITFALLS, AND TRIUMPHS
and x = v; so that
Z √
227
Z
√
x2 dx
.
1 − x2 dx = x 1 − x2 + √
1 − x2
Here we may use a little dodge, for we can write
Z √
Z
Z
Z
(1 − x2 ) dx
dx
x2 dx
2
√
√
1 − x dx =
=
− √
.
1 − x2
1 − x2
1 − x2
Z
x2 dx
√
Adding these two last equations, we get rid of
, and we
1 − x2
have
Z √
Z
√
dx
2
2
2
1 − x dx = x 1 − x + √
.
1 − x2
dx
Do you remember meeting √
? it is got by differentiating
1 − x2
y = arc sin x (see p. 168); hence its integral is arc sin x, and so
√
Z √
x 1 − x2 1
2
1 − x dx =
+ 2 arc sin x + C.
2
You can try now some exercises by yourself; you will find some at
the end of this chapter.
Substitution. This is the same dodge as explained in Chap. IX.,
p. 66. Let
Z us illustrate its application to integration by a few examples.
√
(1)
3 + x dx.
Let
3 + x = u,
Z
replace
1
dx = du;
3
3
u 2 du = 32 u 2 = 32 (3 + x) 2 .
CALCULUS MADE EASY
Z
(2)
dx
.
x + −x
Let
Z
so that
228
dx
x + −x
du
du
= x , and dx = x ;
x = u,
dx
Z
Z
Z
du
du
du
=
=
=
.
1
x (x + −x )
u2 + 1
u u+
u
du
is the result of differentiating arc tan x.
1 + u2
x
Hence
Z tan .
Z
Z the integral is arc
dx
dx
dx
√ .
=
=
(3)
2
2
x + 2x + 3
x + 2x + 1 + 2
(x + 1)2 + ( 2)2
Let
x + 1 = u, dx = du;
Z
du
du
√ ; but 2
then the integral becomes
is the result of
u + a2
u2 + ( 2)2
1
u
differentiating u = arc tan .
a
a
1
x+1
Hence one has finally √ arc tan √ for the value of the given
2
2
integral.
Formulæ of Reduction are special forms applicable chiefly to binomial and trigonometrical expressions that have to be integrated, and
have to be reduced into some form of which the integral is known.
Rationalization, and Factorization of Denominator are dodges applicable in special cases, but they do not admit of any short or general
explanation. Much practice is needed to become familiar with these
preparatory processes.
The following example shows how the process of splitting into partial
fractions, which we learned in Chap. XIII., p. 118, can be made use of
in integration.
DODGES, PITFALLS, AND TRIUMPHS
229
dx
1
;
if
we
split
into partial fracx2 + 2x + 3
x2 + 2x + 3
tions, this becomes (see p. 230):
Z
Z
1
dx
dx
√
√
√
−
2 −2
x + 1 − −2
x + 1 + −2
√
1
x + 1 − −2
√ .
= √
log
2 −2
x + 1 + −2
Notice that the same integral can be expressed sometimes in more than
Z
Take again
one way (which are equivalent to one another).
Pitfalls. A beginner is liable to overlook certain points that a practised hand would avoid; such as the use of factors that are equivalent to
either zero or infinity, and the occurrence of indeterminate quantities
such as 00 . There is no golden rule that will meet every possible case.
Nothing but practice and intelligent care will avail. An example of a
pitfall which had to be circumvented arose in Chap. XVIII., p. 189,
when we came to the problem of integrating x−1 dx.
Triumphs. By triumphs must be understood the successes with
which the calculus has been applied to the solution of problems otherwise intractable. Often in the consideration of physical relations one
is able to build up an expression for the law governing the interaction of the parts or of the forces that govern them, such expression
being naturally in the form of a differential equation, that is an equation containing differential coefficients with or without other algebraic
quantities. And when such a differential equation has been found, one
can get no further until it has been integrated. Generally it is much
easier to state the appropriate differential equation than to solve it:—
the real trouble begins then only when one wants to integrate, unless
CALCULUS MADE EASY
230
indeed the equation is seen to possess some standard form of which the
integral is known, and then the triumph is easy. The equation which
results from integrating a differential equation is called∗ its “solution”;
and it is quite astonishing how in many cases the solution looks as if it
had no relation to the differential equation of which it is the integrated
form. The solution often seems as different from the original expression
as a butterfly does from the caterpillar that it was. Who would have
supposed that such an innocent thing as
1
dy
= 2
dx
a − x2
could blossom out into
y=
1
a+x
log
+ C?
2a
a−x
yet the latter is the solution of the former.
As a last example, let us work out the above together.
By partial fractions,
a2
∗
1
1
1
=
+
,
2
−x
2a(a + x) 2a(a − x)
dx
dx
dy =
+
,
2a(a + x) 2a(a − x)
Z
Z
1
dx
dx
y=
+
2a
a+x
a−x
This means that the actual result of solving it is called its “solution.” But many
mathematicians would say, with Professor Forsyth, “every differential equation is
considered as solved when the value of the dependent variable is expressed as a
function of the independent variable by means either of known functions, or of
integrals, whether the integrations in the latter can or cannot be expressed in terms
of functions already known.”
DODGES, PITFALLS, AND TRIUMPHS
231
1
(log (a + x) − log (a − x))
2a
1
a+x
=
log
+ C.
2a
a−x
=
Not a very difficult metamorphosis!
There are whole treatises, such as Boole’s Differential Equations,
devoted to the subject of thus finding the “solutions” for different original forms.
Exercises XIX.
(1) Find
Z √
Z
(3) Find
−
Z
x2
dx.
a
x log x dx.
Z
(5) Find
1
cos(log x) dx.
x
Z
(log x)a
dx.
x
Z
5x + 1
dx.
2
x +x−2
(7) Find
(9) Find
Z
(11) Find
Z
(13) Find
a2
(See p. 266 for Answers.)
b dx
.
2
x − a2
dx
.
1 − x4
(2) Find
x log x dx.
Z
(4) Find
Z
(6) Find
x2 x dx.
Z
dx
.
x log x
Z
(x2 − 3) dx
.
x3 − 7x + 6
Z
4x dx
.
x4 − 1
Z
dx
.
x a − bx2
(8) Find
(10) Find
(12) Find
(14) Find
x cos x dx.
√
CHAPTER XXI.
FINDING SOME SOLUTIONS.
In this chapter we go to work finding solutions to some important
differential equations, using for this purpose the processes shown in the
preceding chapters.
The beginner, who now knows how easy most of those processes
are in themselves, will here begin to realize that integration is an art.
As in all arts, so in this, facility can be acquired only by diligent and
regular practice. He who would attain that facility must work out
examples, and more examples, and yet more examples, such as are
found abundantly in all the regular treatises on the Calculus. Our
purpose here must be to afford the briefest introduction to serious work.
Example 1. Find the solution of the differential equation
ay + b
Transposing we have
b
dy
= 0.
dx
dy
= −ay.
dx
FINDING SOLUTIONS
233
Now the mere inspection of this relation tells us that we have got
dy
is proportional to y. If we think of the
to do with a case in which
dx
curve which will represent y as a function of x, it will be such that its
slope at any point will be proportional to the ordinate at that point,
and will be a negative slope if y is positive. So obviously the curve
will be a die-away curve (p. 153), and the solution will contain −x as
a factor. But, without presuming on this bit of sagacity, let us go to
work.
As both y and dy occur in the equation and on opposite sides, we
can do nothing until we get both y and dy to one side, and dx to the
other. To do this, we must split our usually inseparable companions
dy and dx from one another.
a
dy
= − dx.
y
b
Having done the deed, we now can see that both sides have got into
1
dy
a shape that is integrable, because we recognize , or dy, as a differy
y
ential that we have met with (p. 143) when differentiating logarithms.
So we may at once write down the instructions to integrate,
Z
Z
dy
a
= − dx;
y
b
and doing the two integrations, we have:
a
log y = − x + log C,
b
where log C is the yet undetermined constant∗ of integration. Then,
∗
We may write down any form of constant as the “constant of integration,” and
the form log C is adopted here by preference, because the other terms in this line
of equation are, or are treated as logarithms; and it saves complications afterward
if the added constant be of the same kind.
CALCULUS MADE EASY
234
delogarizing, we get:
a
y = C− b x ,
which is the solution required. Now, this solution looks quite unlike
the original differential equation from which it was constructed: yet to
an expert mathematician they both convey the same information as to
the way in which y depends on x.
Now, as to the C, its meaning depends on the initial value of y. For
if we put x = 0 in order to see what value y then has, we find that this
makes y = C−0 ; and as −0 = 1 we see that C is nothing else than the
particular value∗ of y at starting. This we may call y0 , and so write the
solution as
a
y = y0 − b x .
Example 2.
Let us take as an example to solve
ay + b
dy
= g,
dx
where g is a constant. Again, inspecting the equation will suggest,
(1) that somehow or other x will come into the solution, and (2) that
if at any part of the curve y becomes either a maximum or a minimum,
dy
g
so that
= 0, then y will have the value = . But let us go to work
dx
a
as before, separating the differentials and trying to transform the thing
∗
Compare what was said about the “constant of integration,” with reference to
Fig. 48 on p. 184, and Fig. 51 on p. 187.
FINDING SOLUTIONS
235
into some integrable shape.
dy
= g − ay;
dx
dy
a g
=
−y ;
dx
b a
dy
a
g = − b dx.
y−
a
b
Now we have done our best to get nothing but y and dy on one
side, and nothing but dx on the other. But is the result on the left side
integrable?
It is of the same form as the result on p. 145; so, writing the instructions to integrate, we have:
Z
Z
a
dy
g = − b dx;
y−
a
and, doing the integration, and adding the appropriate constant,
g
a
log y −
= − x + log C;
a
b
a
g
whence
y − = C− b x ;
a
a
g
and finally,
y = + C− b x ,
a
which is the solution.
If the condition is laid down that y = 0 when x = 0 we can find C;
for then the exponential becomes = 1; and we have
g
+ C,
a
g
C=− .
a
0=
or
CALCULUS MADE EASY
236
Putting in this value, the solution becomes
a
g
y = (1 − − b x ).
a
But further, if x grows indefinitely, y will grow to a maximum; for
g
when x = ∞, the exponential = 0, giving ymax. = . Substituting this,
a
we get finally
a
y = ymax. (1 − − b x ).
This result is also of importance in physical science.
Example 3.
Let
ay + b
dy
= g · sin 2πnt.
dt
We shall find this much less tractable than the preceding. First
divide through by b.
dy a
g
+ y = sin 2πnt.
dt
b
b
Now, as it stands, the left side is not integrable. But it can be
made so by the artifice—and this is where skill and practice suggest a
a
plan—of multiplying all the terms by b t , giving us:
dy a t a a t g a t
b + y b = b · sin 2πnt,
dt
b
b
which is the same as
a
dy a t
d( b t )
g a
b
+y
= b t · sin 2πnt;
dt
dt
b
FINDING SOLUTIONS
237
and this being a perfect differential may be integrated thus:—since, if
a
d( b t )
dy a t
a t du
b
b
= +y
,
u = y ,
dt
dt
dt
Z
at
a
g
y b =
b t · sin 2πnt · dt + C,
b
Z
a
a
g −at
or
y= b
b t · sin 2πnt · dt + C− b t .
[a]
b
The last term is obviously a term which will die out as t increases,
and may be omitted. The trouble now comes in to find the integral
that appears as a factor. To tackle this we resort to the device
(see
Z
p. 224) of integration by parts, the general formula for which is udv =
Z
uv − vdu. For this purpose write

 u = ab t ;
 dv = sin 2πnt · dt.
We shall then have

a
a

 du = b t × dt;
b
1

 v=−
cos 2πnt.
2πn
Inserting these, the integral in question becomes:
Z
a
b t · sin 2πnt · dt
Z
at
a
1
1
a
=−
· b · cos 2πnt − −
cos 2πnt · b t · dt
2πn
2πn
b
Z
a
1 at
a
=−
b cos 2πnt +
b t · cos 2πnt · dt.
2πn
2πnb
[b]
238
CALCULUS MADE EASY
The last integral is still irreducible. To evade the difficulty, repeat
the integration by parts of the left side, but treating it in the reverse
way by writing:

 u = sin 2πnt;
whence
 dv = ab t · dt;


 du = 2πn · cos 2πnt · dt;
b a

 v = bt
a
Inserting these, we get
Z
a
b t · sin 2πnt · dt
b a
2πnb
= · b t · sin 2πnt −
a
a
Z
a
b t · cos 2πnt · dt.
[c]
Noting that the final intractable integral in [c] is the same as that
2πnb
in [b], we may eliminate it, by multiplying [b] by
, and multiplya
a
ing [c] by
, and adding them.
2πnb
The result, when cleared down, is:
Z
at
at
ab · sin 2πnt − 2πnb2 · cos 2πnt
b
b
· sin 2πnt · dt = [d]
a2 + 4π 2 n2 b2
Inserting this value in [a], we get
y=g
a · sin 2πnt − 2πnb · cos 2πnt
a2 + 4π 2 n2 b2
.
To simplify still further, let us imagine an angle φ such that tan φ =
FINDING SOLUTIONS
239
2πnb
.
a
Then
and
2πnb
,
+ 4π 2 n2 b2
a
.
cos φ = √
2
a + 4π 2 n2 b2
sin φ = √
a2
Substituting these, we get:
y=g
cos φ · sin 2πnt − sin φ · cos 2πnt
√
,
a2 + 4π 2 n2 b2
which may be written
sin(2πnt − φ)
y = g√
,
a2 + 4π 2 n2 b2
which is the solution desired.
This is indeed none other than the equation of an alternating electric
current, where g represents the amplitude of the electromotive force,
n the frequency, a the resistance, b the coefficient of self-induction of
the circuit, and φ is an angle of lag.
Example 4.
Suppose that
M dx + N dy = 0.
We could integrate this expression directly, if M were a function of x
only, and N a function of y only; but, if both M and N are functions
that depend on both x and y, how are we to integrate it? Is it itself
an exact differential? That is: have M and N each been formed by
CALCULUS MADE EASY
240
partial differentiation from some common function U , or not? If they
have, then

∂U


= M,

∂x
∂U


= N.

∂y
And if such a common function exists, then
∂U
∂U
dx +
dy
∂x
∂y
is an exact differential (compare p. 172).
Now the test of the matter is this. If the expression is an exact
differential, it must be true that
dM
dN
=
;
dy
dx
d(dU )
d(dU )
=
,
dx dy
dy dx
for then
which is necessarily true.
Take as an illustration the equation
(1 + 3xy) dx + x2 dy = 0.
Is this an exact differential or not? Apply the test.

d(1 + 3xy)


= 3x,

dy
2


 d(x )
= 2x,
dx
which do not agree. Therefore, it is not an exact differential, and the
two functions 1 + 3xy and x2 have not come from a common original
function.
FINDING SOLUTIONS
241
It is possible in such cases to discover, however, an integrating factor, that is to say, a factor such that if both are multiplied by this factor,
the expression will become an exact differential. There is no one rule
for discovering such an integrating factor; but experience will usually
suggest one. In the present instance 2x will act as such. Multiplying
by 2x, we get
(2x + 6x2 y) dx + 2x3 dy = 0.
Now apply the test to this.

d(2x + 6x2 y)


= 6x2 ,

dy

d(2x3 )


= 6x2 ,
dx
which agrees. Hence this is an exact differential, and may be integrated.
Now, if w = 2x3 y,
Z
Hence
dw = 6x2 y dx + 2x3 dy.
Z
2
6x y dx + 2x3 dy = w = 2x3 y;
so that we get
U = x2 + 2x3 y + C.
d2 y
+ n2 y = 0.
dt2
In this case we have a differential equation of the second degree, in
Example 5. Let
which y appears in the form of a second differential coefficient, as well
as in person.
Transposing, we have
d2 y
= −n2 y.
dt2
242
CALCULUS MADE EASY
It appears from this that we have to do with a function such that its
second differential coefficient is proportional to itself, but with reversed
sign. In Chapter XV. we found that there was such a function—namely,
the sine (or the cosine also) which possessed this property. So, without
further ado, we may infer that the solution will be of the form y =
A sin(pt + q). However, let us go to work.
Multiply both sides of the original equation by 2
dy
and integrate,
dt
d2 y dy
dy
+ 2x2 y
= 0, and, as
2
dt dt
dt
2
dy
2
d
2
d y dy
dy
dt
2 2
=
,
+ n2 (y 2 − C 2 ) = 0,
dt dt
dt
dt
giving us 2
C being a constant. Then, taking the square roots,
p
dy
dy
= −n y 2 − C 2 and p
= n · dt.
dt
C 2 − y2
But it can be shown that (see p. 168)
1
d(arc sin
p
=
C 2 − y2
dy
y
)
C ;
whence, passing from angles to sines,
arc sin
y
= nt + C1
C
and y = C sin(nt + C1 ),
where C1 is a constant angle that comes in by integration.
Or, preferably, this may be written
y = A sin nt + B cos nt, which is the solution.
243
FINDING SOLUTIONS
d2 y
− n2 y = 0.
dt2
Here we have obviously to deal with a function y which is such
Example 6.
that its second differential coefficient is proportional to itself. The only
function we know that has this property is the exponential function
(see p. 139), and we may be certain therefore that the solution of the
equation will be of that form.
Proceeding as before, by multiplying through by 2
dy
, and integratdx
dy
d2 y dy
− 2x2 y
= 0,
2
dx dx
dx
2
dy
2
d
2
dy
d y dy
dx
and, as 2 2
=
,
− n2 (y 2 + c2 ) = 0,
dx dx
dx
dx
p
dy
− n y 2 + c2 = 0,
dx
dy
where c is a constant, and p
= n dx.
2 + c2
y
p
Now, if w = log (y + y 2 + c2 ) = log u,
p
y + y 2 + c2
dw
1
du
y
= ,
=1+ p
= p
du
u
dy
y 2 + c2
y 2 + c2
dw
1
.
and
=p
dy
y 2 + c2
ing, we get 2
Hence, integrating, this gives us
p
log (y + y 2 + c2 ) = nx + log C,
p
y + y 2 + c2 = Cnx .
p
p
Now
(y + y 2 + c2 ) × (−y + y 2 + c2 ) = c2 ;
p
c2
whence
−y + y 2 + c2 = −nx .
C
(1)
(2)
CALCULUS MADE EASY
244
Subtracting (2) from (1) and dividing by 2, we then have
1
1 c2 −nx
y = Cnx −
,
2
2C
which is more conveniently written
y = Anx + B−nx .
Or, the solution, which at first sight does not look as if it had anything
to do with the original equation, shows that y consists of two terms,
one of which grows logarithmically as x increases, and of a second term
which dies away as x increases.
Example 7.
dy
d2 y
+ a + gy = 0.
2
dt
dt
Examination of this expression will show that, if b = 0, it has the
Let
b
form of Example 1, the solution of which was a negative exponential.
On the other hand, if a = 0, its form becomes the same as that of
Example 6, the solution of which is the sum of a positive and a negative
exponential. It is therefore not very surprising to find that the solution
of the present example is
y = (−mt )(Ant + B−nt ),
r
a
a2
g
where
m=
and n =
− .
2
2b
4b
b
The steps by which this solution is reached are not given here; they
may be found in advanced treatises.
FINDING SOLUTIONS
245
Example 8.
2
d2 y
2d y
=
a
.
dt2
dx2
It was seen (p. 174) that this equation was derived from the original
y = F (x + at) + f (x − at),
where F and f were any arbitrary functions of t.
Another way of dealing with it is to transform it by a change of
variables into
d2 y
= 0,
du · dv
where u = x + at, and v = x − at, leading to the same general solution.
If we consider a case in which F vanishes, then we have simply
y = f (x − at);
and this merely states that, at the time t = 0, y is a particular function of x, and may be looked upon as denoting that the curve of the
relation of y to x has a particular shape. Then any change in the value
of t is equivalent simply to an alteration in the origin from which x is
reckoned. That is to say, it indicates that, the form of the function
being conserved, it is propagated along the x direction with a uniform
velocity a; so that whatever the value of the ordinate y at any particular time t0 at any particular point x0 , the same value of y will
appear at the subsequent time t1 at a point further along, the abscissa
of which is x0 +a(t1 −t0 ). In this case the simplified equation represents
the propagation of a wave (of any form) at a uniform speed along the
x direction.
CALCULUS MADE EASY
246
If the differential equation had been written
m
d2 y
d2 y
=
k
,
dt2
dx2
the solution would have been the same, but the velocity of propagation
would have had the value
r
a=
k
.
m
You have now been personally conducted over the frontiers into the
enchanted land. And in order that you may have a handy reference
to the principal results, the author, in bidding you farewell, begs to
present you with a passport in the shape of a convenient collection of
standard forms (see pp. 249–251). In the middle column are set down
a number of the functions which most commonly occur. The results of
differentiating them are set down on the left; the results of integrating
them are set down on the right. May you find them useful!
EPILOGUE AND APOLOGUE.
It may be confidently assumed that when this tractate “Calculus made
Easy” falls into the hands of the professional mathematicians, they will
(if not too lazy) rise up as one man, and damn it as being a thoroughly
bad book. Of that there can be, from their point of view, no possible manner of doubt whatever. It commits several most grievous and
deplorable errors.
First, it shows how ridiculously easy most of the operations of the
calculus really are.
Secondly, it gives away so many trade secrets. By showing you that
what one fool can do, other fools can do also, it lets you see that these
mathematical swells, who pride themselves on having mastered such an
awfully difficult subject as the calculus, have no such great reason to
be puffed up. They like you to think how terribly difficult it is, and
don’t want that superstition to be rudely dissipated.
Thirdly, among the dreadful things they will say about “So Easy”
is this: that there is an utter failure on the part of the author to
demonstrate with rigid and satisfactory completeness the validity of
sundry methods which he has presented in simple fashion, and has
even dared to use in solving problems! But why should he not? You
don’t forbid the use of a watch to every person who does not know how
CALCULUS MADE EASY
248
to make one? You don’t object to the musician playing on a violin that
he has not himself constructed. You don’t teach the rules of syntax to
children until they have already become fluent in the use of speech. It
would be equally absurd to require general rigid demonstrations to be
expounded to beginners in the calculus.
One other thing will the professed mathematicians say about this
thoroughly bad and vicious book: that the reason why it is so easy is
because the author has left out all the things that are really difficult.
And the ghastly fact about this accusation is that—it is true! That
is, indeed, why the book has been written—written for the legion of
innocents who have hitherto been deterred from acquiring the elements
of the calculus by the stupid way in which its teaching is almost always presented. Any subject can be made repulsive by presenting it
bristling with difficulties. The aim of this book is to enable beginners
to learn its language, to acquire familiarity with its endearing simplicities, and to grasp its powerful methods of solving problems, without being compelled to toil through the intricate out-of-the-way (and
mostly irrelevant) mathematical gymnastics so dear to the unpractical
mathematician.
There are amongst young engineers a number on whose ears the
adage that what one fool can do, another can, may fall with a familiar
sound. They are earnestly requested not to give the author away, nor
to tell the mathematicians what a fool he really is.
TABLE OF STANDARD FORMS.
dy
dx
←−
y
Z
−→
y dx
Algebraic.
1
x
1 2
x
2
0
a
ax + C
1
x±a
a
ax
1
ax2
2
2x
x2
1 3
x
3
nxn−1
xn
1
xn+1 + C
n+1
−x−2
x−1
log x + C
du dv dw
±
±
dx dx
dx
dv
du
u
+v
dx
dx
du
dv
v
−u
dx
dx
v2
du
dx
u±v±w
1 2
x
2
R
u dx ±
+C
± ax + C
R
+C
+C
v dx ±
R
w dx
uv
No general form known
u
v
No general form known
u
ux −
R
x du + C
250
CALCULUS MADE EASY
dy
dx
←−
Z
−→
y
y dx
Exponential and Logarithmic.
x
x
x + C
x−1
log x
0.4343 × x−1
log10 x
x(log x − 1) + C
ax log a
0.4343x(log x − 1) + C
ax
ax
+C
log a
Trigonometrical.
cos x
sin x
− sin x
cos x
sec2 x
− cos x + C
sin x + C
− log cos x + C
tan x
Circular (Inverse).
1
p
(1 − x2 )
1
−p
(1 − x2 )
1
1 + x2
arc sin x
x · arc sin x +
√
1 − x2 + C
arc cos x
x · arc cos x −
√
1 − x2 + C
arc tan x
x · arc tan x − 12 log (1 + x2 ) + C
Hyperbolic.
cosh x
sinh x
cosh x + C
sinh x
cosh x
sinh x + C
sech2 x
tanh x
log cosh x + C
251
TABLE OF STANDARD FORMS
dy
dx
←−
Z
−→
y
y dx
Miscellaneous.
−
−
1
(x + a)2
x
3
2
(a2 + x2 )
b
∓
(a ± bx)2
3a2 x
−
5
(a2 + x2 ) 2
1
x+a
1
√
2
a + x2
3
−a · sin ax
cos ax
a · sec2 ax
tan ax
sin 2x
sin2 x
− sin 2x
cos2 x
n · sinn−1 x · cos x
sinn x
cos x
sin2 x
sin 2x
− 4
sin x
sin2 x − cos2 x
sin2 x · cos2 x
1
sin x
1
sin2 x
1
sin x · cos x
n · sin mx · cos nx +
sin mx · sin nx
2a · sin 2ax
sin2 ax
−2a · sin 2ax
cos2 ax
√
a2 + x2 ) + C
1
± log (a ± bx) + C
b
x
√
+C
2
a + x2
(a2 + x2 ) 2
sin ax
m · sin nx · cos mx
log (x +
1
a ± bx
a2
a · cos ax
−
log (x + a) + C
1
− cos ax + C
a
1
sin ax + C
a
1
− log cos ax + C
a
x sin 2x
−
+C
2
4
x sin 2x
+
+C
2
4
−
n−1
cos x
sinn−1 x +
n
n
log tan
Z
sinn−2 x dx + C
x
+C
2
− cotan x + C
log tan x + C
1
2
cos(m − n)x − 12 cos(m + n)x + C
x sin 2ax
−
+C
2
4a
x sin 2ax
+
+C
2
4a
ANSWERS.
(p. 24.)
Exercises I.
(1)
dy
= 13x12 .
dx
(2)
dy
3 5
= − x− 2 .
dx
2
(3)
dy
= 2ax(2a−1) .
dx
(4)
du
= 2.4t1.4 .
dt
(5)
dz
1 2
= u− 3 .
du
3
(6)
dy
5 8
= − x− 3 .
dx
3
(7)
8 13
du
= − x− 5 .
dx
5
(8)
dy
= 2axa−1 .
dx
(9)
dy
3 3−q
= x q .
dx
q
(10)
dy
m m+n
= − x− n .
dx
n
Exercises II.
(p. 31.)
(1)
dy
= 3ax2 .
dx
(2)
dy
1
= 13 × 32 x 2 .
dx
(3)
dy
1
= 6x− 2 .
dx
(4)
dy
1 1 1
= c 2 x− 2 .
dx
2
(5)
du
an n−1
=
z .
dz
c
(6)
dy
= 2.36t.
dt
dlt
= 0.000012 × l0 .
dt
dC
(8)
= abV b−1 , 0.98, 3.00 and 7.47 candle power per volt respectively.
dV
r
r
dn
1
gT
dn
1
gT
(9)
=−
,
=−
,
2
2
dD
LD rπσ
dL
DL
πσ
r
dn
1
gT
dn
1
g
=−
,
=
.
3
dσ
2DL πσ
dT
2DL πσT
(7)
253
ANSWERS TO EXERCISES
D
Rate of change of P when t varies
=− .
Rate of change of P when D varies
t
dD
0.000012lt
(11) 2π, 2πr, πl, 23 πrh, 8πr, 4πr2 .
(12)
=
.
dT
π
(10)
Exercises III.
x2 x3 x4
(1) (a) 1 + x +
+
+
+ ...
2
6
24
(d ) 3x2 + 6ax + 3a2 .
(2)
dw
= a − bt.
dt
(3)
(p. 45.)
(b) 2ax + b.
(c) 2x + 2a.
dy
= 2x.
dx
(4) 14110x4 − 65404x3 − 2244x2 + 8192x + 1379.
(5)
dx
= 2y + 8.
dy
(6) 185.9022654x2 + 154.36334.
(7)
−5
.
(3x + 2)2
(8)
6x4 + 6x3 + 9x2
.
(1 + x + 2x2 )2
(9)
ad − bc
.
(cx + d)2
(10)
anx−n−1 + bnxn−1 + 2nx−1
.
(x−n + b)2
(11) b + 2ct.
b
R0 (a + 2bt)
(12) R0 (a+2bt), R0 a + √ , −
(1 + at + bt2 )2
2 t
or
R2 (a + 2bt)
.
R0
(13) 1.4340(0.000014t − 0.001024), −0.00117, −0.00107, −0.00097.
(14)
dE
k
=b+ ,
dl
i
dE
c + kl
=− 2 .
di
i
254
CALCULUS MADE EASY
Exercises IV.
(1) 17 + 24x; 24.
(2)
(p. 50.)
x2 + 2ax − a
;
(x + a)2
2a(a + 1)
.
(x + a)3
x2
x3
x2
+
; 1+x+
.
1×2 1×2×3
1×2
(4) (Exercises III.):
(3) 1 + x +
d3 y
d2 y
=
= 1 + x + 12 x2 + 16 x3 + . . ..
dx2
dx3
(b) 2a, 0.
(c) 2, 0.
(d ) 6x + 6a, 6.
(1) (a)
(2) −b, 0.
(3) 2, 0.
(4) 56440x3 − 196212x2 − 4488x + 8192.
169320x2 − 392424x − 4488.
(5) 2, 0.
(6) 371.80453x, 371.80453.
30
270
(7)
, −
.
3
(3x + 2)
(3x + 2)4
(Examples, p. 40):
6a
6a
(1) 2 x,
.
b
b2
2
1.056
(3) √
− √
,
3 8
5 11
θ
θ
√
√
3a b 6b 3 a
(2) √ −
,
x3
2 x
2.3232
16
√
− √
.
5 16
3 11
θ
3 θ
√
√
18b 3 a 3a b
− √ .
x4
4 x3
(4) 810t4 − 648t3 + 479.52t2 − 139.968t + 26.64.
3240t3 − 1944t2 + 959.04t − 139.968.
(5) 12x + 2, 12.
(6) 6x2 − 9x, 12x − 9.
3
1
1
1 15
1
√ +√
√ −√
(7)
+
.
5
7
3
4
4
θ
θ
θ
θ
3
1
1
15
7
1
√ −√
√ +√
−
.
8
8
θ5
θ3
θ9
θ7
ANSWERS TO EXERCISES
Exercises V.
255
(p. 63.)
(2) 64; 147.2; and 0.32 feet per second.
(3) x = a − gt; ẍ = −g.
(4) 45.1 feet per second.
(5) 12.4 feet per second per second. Yes.
(6) Angular velocity = 11.2 radians per second; angular acceleration
= 9.6 radians per second per second.
(7) v = 20.4t2 − 10.8. a = 40.8t. 172.8 in./sec., 122.4 in./sec2 .
1
(8) v =
30
p
3
(t −
(9) v = 0.8 −
125)2
, a=−
1
p
.
3
45 (t − 125)5
8t
24t2 − 32
,
a
=
, 0.7926 and 0.00211.
(4 + t2 )2
(4 + t2 )3
(10) n = 2, n = 11.
Exercises VI.
(1) √
x
.
+1
(2) √
x2
(4) p
ax
(a − x2 )3
.
x
.
+ a2
x2
(p. 72.)
1
(3) − p
.
2 (a + x)3
2a2 − x2
√
(5)
.
x 3 x 2 − a2
CALCULUS MADE EASY
(6)
(8)
3 2
x
2
(7)
2a (x − a)
.
(x + a)3
(9)
1
√
.
(1 − θ) 1 − θ2
8
3
4
x
(x
+
a)
−
(x
+
a)
9
2
3
(x4 + a) 3 (x3 + a) 2
5 3
y .
2
Exercises VII.
(1)
256
(p. 74.)
3x2 (3 + 3x3 )
dw
=
3 .
dx
27 12 x3 + 14 x6
dv
12x
= −p
√
2 .
p
√
√
dx
2
2
1 + 2 + 3x
3 + 4 1 + 2 + 3x
√
x2 3 + x3
du
(3)
= − v"
u
dx
2 #3
3
u
x
t 1+ 1+ √
3
(2)
Exercises VIII.
(p. 88.)
(2) 1.44.
dy
= 3x2 + 3; and the numerical values are: 3, 3 34 , 6, and 15.
dx
√
(5) ± 2.
(4)
(6)
dy
4x
1
= − . Slope is zero where x = 0; and is ∓ √ where x = 1.
dx
9y
3 2
ANSWERS TO EXERCISES
(7) m = 4, n = −3.
(8) Intersections at x = 1, x = −3. Angles 153◦ 260 , 2◦ 280 .
(9) Intersection at x = 3.57, y = 3.50. Angle 16◦ 160 .
(10) x = 13 , y = 2 13 , b = − 35 .
Exercises IX.
(p. 107.)
(1) Min.: x = 0, y = 0; max.: x = −2, y = −4.
(2) x = a.
(5)
√
(4) 25 3 square inches.
dy
10
10
=− 2 +
; x = 4; y = 5.
dx
x
(8 − x)2
(6) Max. for x = −1; min. for x = 1.
(7) Join the middle points of the four sides.
R
(8) r = 23 R, r = , no max.
2
r
2
R
(9) r = R
, r = √ , r = 0.8506R.
3
2
8
(10) At the rate of square feet per second.
r
√
r
R 8
NR
(11) r =
.
(12) n =
.
3
r
257
CALCULUS MADE EASY
Exercises X.
258
(p. 115.)
(1) Max.: x = −2.19, y = 24.19; min.:, x = 1.52, y = −1.38.
(2)
b
d2 y
b
dy
= − 2cx; 2 = −2c; x =
(a maximum).
dx
a
dx
2ac
(3) (a) One maximum and two minima.
(b) One maximum. (x = 0; other points unreal.)
(4) Min.: x = 1.71, y = 6.14.
(5) Max: x = −.5, y = 4.
(6) Max.: x = 1.414, y = 1.7675.
Min.: x = −1.414, y = 1.7675.
(7) Max.: x = −3.565, y = 2.12.
Min.: x = +3.565, y = 7.88.
(8) 0.4N , 0.6N .
r
(9) x =
a
.
c
(10) Speed 8.66 nautical miles per hour. Time taken 115.47 hours.
Minimum cost £112. 12s.
(11) Max. and min. for x = 7.5, y = ±5.414. (See example no. 10,
p. 71.)
(12) Min.: x = 12 , y = 0.25; max.: x = − 31 , y = 1.408.
ANSWERS TO EXERCISES
Exercises XI.
(1)
(4)
1
2
+
.
x−3 x+4
(2)
5
4
−
.
x−4 x−3
(p. 127.)
1
2
+
.
x−1 x−2
(5)
259
(3)
2
1
+
.
x−3 x+4
19
22
−
.
13(2x + 3) 13(3x − 2)
2
4
5
+
−
.
x−2 x−3 x−4
1
11
1
(7)
+
+
.
6(x − 1) 15(x + 2) 10(x − 3)
(6)
7
71
5
+
−
.
9(3x + 1) 63(3x − 2) 7(2x + 1)
1
2x + 1
2
1 − 2x
(9)
+
.
(10)
x
+
+
.
3(x − 1) 3(x2 + x + 1)
3(x + 1) 3(x2 − x + 1)
(8)
(11)
2x + 1
3
+ 2
.
(x + 1) x + x + 1
(13)
1
1
1
−
+
.
4(x − 1) 4(x + 1) 2(x + 1)2
(14)
(12)
1
1
2
−
+
.
x − 1 x − 2 (x − 2)2
4
4
1
.
−
−
9(x − 1) 9(x + 2) 3(x + 2)2
(15)
1
x−1
1
− 2
− 2
.
x + 2 x + x + 1 (x + x + 1)2
(16)
5
32
36
−
+
.
2
x + 4 (x + 4)
(x + 4)3
(17)
7
55
73
+
+
.
9(3x − 2)2 9(3x − 2)3 9(3x − 2)4
(18)
x
1
1
+
−
.
2
2
6(x − 2) 3(x − 2)
6(x + 2x + 4)
260
CALCULUS MADE EASY
(p. 150.)
Exercises XII.
(1) ab(ax + −ax ).
2
(2) 2at + .
t
(5) npv n−1 .
(6)
n
.
x
x
3− x−1
(7)
.
(x − 1)2
(8) 6x−5x − 5(3x2 + 1)−5x .
(10)
(11)
1
6x
+ √
2
3x − 1 2 ( x + x)
(9)
axa−1
.
xa + a
√
(3x2 − 1) ( x + 1).
1 − log (x + 3)
.
(x + 3)2
(12) ax (axa−1 + xa log a).
(15)
(3) log n.
1+x
.
x
(14) Min.: y = 0.7 for x = 0.694.
(16)
Exercises XIII.
(1) Let
3
(log ax)2 .
x
(p. 160.)
t
= x (∴ t = 8x), and use the Table on page 157.
T
(2) T = 34.627; 159.46 minutes.
(3) Take 2t = x; and use the Table on page 157.
x
(5) (a) xx (1 + log x); (b) 2x(x )x ; (c) x × xx (1 + log x).
ANSWERS TO EXERCISES
(6) 0.14 second.
(7) (a) 1.642; (b) 15.58.
(8) µ = 0.00037, 31m 14 .
(9) i is 63.4% of i0 , 220 kilometres.
(10) 0.133, 0.145, 0.155, mean 0.144; −10.2%, −0.9%, +77.2%.
1
(11) Min. for x = .
(12) Max. for x = .
(13) Min. for x = log a.
Exercises XIV.
(p. 170.)
dy
π
;
= A cos θ −
dθ
2
dy
dy
(ii)
= 2 sin θ cos θ = sin 2θ and
= 2 cos 2θ;
dθ
dθ
dy
dy
(iii)
= 3 sin2 θ cos θ and
= 3 cos 3θ.
dθ
dθ
π
dy
(2) θ = 45◦ or radians.
(3)
= −n sin 2πnt.
4
dt
(1) (i)
(4) ax log a cos ax .
(6) 18.2 cos (x + 26◦ ).
(5)
cos x
= cotan x
sin x
261
262
CALCULUS MADE EASY
dy
= 100 cos (θ − 15◦ ), which is a maximum when
dθ
(θ −15◦ ) = 0, or θ = 15◦ ; the value of the slope being then = 100.
(7) The slope is
When θ = 75◦ the slope is 100 cos(75◦ − 15◦ ) = 100 cos 60◦ =
100 ×
1
2
= 50.
(8) cos θ sin 2θ + 2 cos 2θ sin θ = 2 sin θ cos2 θ + cos 2θ
= 2 sin θ 3 cos2 θ − 1 .
(9) amnθn−1 tanm−1 (θn ) sec2 θn .
(10) x sin2 x + sin 2x ; x sin2 x + 2 sin 2x + 2 cos 2x .
(11) (i)
(12) (i)
dy
ab
a −x
1◦
ab
=
b ; (iii)
.
× 2
2 ; (ii)
dx
b
90
(b + x2 )
(x + b)
dy
= sec x tan x;
dx
1
dy
= −√
(ii)
;
dx
1 − x2
dy
1
=
;
(iii)
dx
1 + x2
dy
1
(iv)
= √
;
dx
x x2 − 1
√
3 sec x (3 sec2 x − 1)
dy
(v)
=
.
dx
2
dy
= 4.6 (2θ + 3)1.3 cos (2θ + 3)2.3 .
dθ
dy
(14)
= 3θ2 + 3 cos (θ + 3) − log 3 cos θ × 3sin θ + 3θ .
dθ
(13)
(15) θ = cot θ; θ = ±0.86; is max. for +θ, min. for −θ.
263
ANSWERS TO EXERCISES
Exercises XV.
(1) x3 − 6x2 y − 2y 2 ;
1
3
(p. 177.)
− 2x3 − 4xy.
(2) 2xyz + y 2 z + z 2 y + 2xy 2 z 2 ;
2xyz + x2 z + xz 2 + 2x2 yz 2 ;
2xyz + x2 y + xy 2 + 2x2 y 2 z.
(3)
(x + y + z) − (a + b + c) 3
1
{(x − a) + (y − b) + (z − c)} =
; .
r
r
r
(4) dy = vuv−1 du + uv log u dv.
(5) dy = 3 sin vu2 du + u3 cos v dv,
dy = u sin xu−1 cos x dx + (sin x)u log sin xdu,
1
1 1
du − log u 2 dv.
dy =
v u
v
(7) Minimum for x = y = − 12 .
(8) (a) Length 2 feet, width = depth = 1 foot, vol. = 2 cubic feet.
2
(b) Radius = feet = 7.46 in., length = 2 feet, vol. = 2.54.
π
(9) All three parts equal; the product is maximum.
(10) Minimum for x = y = 1.
(11) Min.: x =
1
2
and y = 2.
(12) Angle at apex = 90◦ ; equal sides = length =
√
3
2V .
264
CALCULUS MADE EASY
(p. 187.)
Exercises XVI.
(1) 1 31 .
(2) 0.6344.
(3) 0.2624.
(4) (a) y = 18 x2 + C; (b) y = sin x + C.
(5) y = x2 + 3x + C.
Exercises XVII.
√ 3
4 ax 2
+ C.
(1)
3
(4)
1 3
x
3
(2) −
+ ax + C.
(6) x4 + x3 + x2 + x + C.
(8)
(9)
1
+ C.
x3
(3)
x4
+ C.
4a
5
(5) −2x− 2 + C.
ax2 bx3 cx4
(7)
+
+
+ C.
4
9
16
x2 + a
a2 + a
= x−a+
by division. Therefore the answer is
x+a
x+a
x2
− ax + (a2 + a) log (x + a) + C. (See pages 196 and 198.)
2
x4
27
x3 2 − a 2
+ 3x3 + x2 + 27x + C. (10)
+
x − 2ax + C.
4
2
3
2
3
4
(11) a2 (2x 2 + 49 x 3 ) + C.
(13)
(p. 202.)
θ sin 2aθ
+
+ C.
2
4a
(12) − 13 cos θ − 16 θ + C.
(14)
θ sin 2θ
−
+ C.
2
4
ANSWERS TO EXERCISES
(15)
θ sin 2aθ
−
+ C.
2
4a
(16)
265
1 3x
.
3
(18) − log (1 − x) + C.
(17) log(1 + x) + C.
Exercises XVIII.
(p. 221.)
(1) Area = 60; mean ordinate = 10.
√
(2) Area = 23 of a × 2a a.
2
= 0.637.
π
(4) Area = 1.57; mean ordinate = 0.5.
(3) Area = 2; mean ordinate =
(5) 0.572, 0.0476.
h
(6) Volume = πr2 .
3
(7) 1.25.
(8) 79.4.
(9) Volume = 4.9348; area of surface = 12.57 (from 0 to π).
a
(10) a log a,
log a.
a−1
(12) Arithmetical mean = 9.5; quadratic mean = 10.85.
1 p 2
(13) Quadratic mean = √
A1 + A23 ; arithmetical mean = 0.
2
The first involves a somewhat difficult integral, and may be
stated thus: By definition the quadratic mean will be
s
Z 2π
1
(A1 sin x + A3 sin 3x)2 dx.
2π 0
CALCULUS MADE EASY
266
Now the integration indicated by
Z
(A21 sin2 x + 2A1 A3 sin x sin 3x + A23 sin2 3x) dx
is more readily obtained if for sin2 x we write
1 − cos 2x
.
2
For 2 sin x sin 3x we write cos 2x − cos 4x; and, for sin2 3x,
1 − cos 6x
.
2
Making these substitutions, and integrating, we get (see p. 198)
A21
sin 2x
A23
sin 6x
sin 2x sin 4x
x−
+ A1 A3
−
+
x−
.
2
2
2
4
2
6
At the lower limit the substitution of 0 for x causes all this to
vanish, whilst at the upper limit the substitution of 2π for x gives
A21 π + A23 π. And hence the answer follows.
(14) Area is 62.6 square units. Mean ordinate is 10.42.
(16) 436.3. (This solid is pear shaped.)
Exercises XIX.
(p. 231.)
√
x2
x a2 − x 2 a2
x
(log x − 12 ) + C.
(1)
+ sin−1 + C. (2)
2
2
2
a
ANSWERS TO EXERCISES
xa+1
(3)
a+1
1
log x −
a+1
(5) sin(log x) + C.
(7)
1
(log x)a+1 + C.
a+1
267
x
+ C. (4) sin + C.
(6) x (x2 − 2x + 2) + C.
(8) log (log x) + C.
(9) 2 log (x − 1) + 3 log (x + 2) + C.
(10)
1
2
log (x − 1) + 15 log (x − 2) +
(11)
b
x−a
log
+ C.
2a
x+a
3
10
log (x + 3) + C.
(12) log
x2 − 1
+ C.
x2 + 1
1+x 1
+ arc tan x + C.
1−x 2
√
√
1
a − a − bx2
1
√
(14) √ log
. (Let
= v; then, in the result, let
x
x a
ra
b
v 2 − = v − u.)
a
You had better differentiate now the answer and work back to
(13)
1
4
log
the given expression as a check.
Every earnest student is exhorted to manufacture more examples
for himself at every stage, so as to test his powers. When integrating
he can always test his answer by differentiating it, to see whether he
gets back the expression from which he started.
There are lots of books which give examples for practice. It will suffice here to name two: R. G. Blaine’s The Calculus and its Applications,
and F. M. Saxelby’s A Course in Practical Mathematics.
GLASGOW: PRINTED AT THE UNIVERSITY PRESS BY ROBERT MACLEHOSE AND CO. LTD.
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transcriber’s note
The diagrams have been re-created, using accompanying
formulas or descriptions from the text where possible.
In Chapter XIV, pages 132–159, numerical values of 1 +
1 n
n ,
x , and related quantities of British currency have been verified
and rounded to the nearest digit.
On page 142 (page 146 in the original), the graphs of the
natural logarithm and exponential functions, Figures 38 and 39,
have been interchanged to match the surrounding text.
The vertical dashed lines in the natural logarithm graph,
Figure 39 (Figure 38 in the original), have been moved to
match the data in the corresponding table.
On page 164 (page 167 in the original), the graphs of the sine
and cosine functions, Figures 44 and 45, have been interchanged
to match the surrounding text.
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